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WORKS  OF  C.  E.  GREENE 

PUBLISHED    BY 

JOHN  WILEY  &.SONS. 


Graphics  for  Engineers,  Architects,  and  Builders. 

A  manual  for  designers,  and  a  text-book  for  scientific 
schools. 

Trusses  and  Arches : 

Analyzed  and  Discussed  by  Graphical  Methods.  In 
three  parts — published  separately. 

Part  I.    Roof  Trusses  : 

Diagrams  for  Steady  Load,  Snow  and  Wind.  8vo,  80 
pp.,  3  folding  plates.  Revised  Edition.  $1.25. 

Part  II.    Bridge  Trusses: 

Single,  Continuous,  and  Draw  Spans;  Single  and  Mul- 
tiple Systems;  Straight  and  Inclined  Chords.  8vo,  190 
pp.,  10  folding  plates.  Fifth  Edition,  Revised.  $2.50. 

Part  III.    Arches, 

In  Wood,  Iron,  and  Stone,  for  Roofs,  Bridges,  and  \\  all- 
openings  ;  Arched  Ribs  and  Braced  Arches;  Stresses 
from  Wind  and  Change  of  Temperature.  8vo,  194  pp., 
8  folding  plates.  Third  Edition,  Revised.  $2.50. 

Structural  Mechanics. 

Comprising  the  Strengtn  and  Resistance  of  Materials 
and  Elements  of  Structural  Design.  With  Examples 
and  Problems.  By  the  late  Charles  E.  Greene,  A.M., 
C.E.  New  Edition,  Revised  and  Enlarged  by  A.  E. 
Greene.  8vo,  viii  -f-  244  pages,  99  figures.  $2.50  net. 


STRUCTURAL  MECHANICS 


COMPRISING   THE 


STRENGTH    AND    RESISTANCE    OF    MATERIALS    AND 
ELEMENTS    OF    STRUCTURAL    DESIGN 


WITH  EXAMPLES  AND   PROBLEMS 


BY 

CHARLES   E.  GREENE,  A.M.,   C.E. 

LATE    PROFESSOR    OF     CIVIL     ENGINEERING,     UNIVERSITY     OF'  MICHIGAN 


REVISED    BY    A.    E.    GREENE 

JUNIOR    PROFESSOR   OF   CIVIL   ENGINEERING,    UNIVERSITY   OF   MICHIGAN 


THIRD  EDITION,  REVISED 

FIRST    THOUSAND 


NEW  YORK 

JOHN    WILEY    &    SONS 

LONDON:    CHAPMAN   &    HALL,    LIMITED 

1909. 


Copyright,  1897, 

BY 
CHARLES  E.  GREENB 

Copyright,  1905, 

BY 

A.  E.  GREENE 


If*  Scientific  |Jresa 
^Bruntmonb  anb  Company 


PREFACE. 


THE  author,  in  teaching  for  many  years  the  subjects  embraced 
in  the  following  pages,  has  found  it  advantageous  to  take  at 
first  but  a  portion  of  what  is  included  in  the  several  chapters, 
and,  after  a  general  survey  of  the  field,  to  return  and  extend  the 
investigation  more  in  detail.  Some  of  the  sections,  therefore, 
are  not  leaded  and  can  be  omitted  at  first  reading.  A  few  of 
the  special  investigations  may  become  of  interest  only  when  the 
problems  to  which  they  relate  occur  in  actual  practice. 

It  is  hoped  that  this  book  will  be  serviceable  after  the  class- 
room work  is  concluded,  and  reference  is  facilitated  by  a  more 
compact  arrangement  of  the  several  matters  than  the  course 
suggested  above  would  give.  The  attempt  has  been  made  to 
deal  with  practicable  cases,  and  the  examples  for  the  most  part 
are  shaped  with  that  end  in  view.  A  full  index  will  enable  one 
to  find  any  desired  topic. 

The  treatment  of  the  subject  of  internal  stress  is  largely 
graphical.  All  the  constructions  are  simple,  and  the  results, 
besides  being  useful  in  themselves,  shed  much  light  on  various 
problems.  The  time  devoted  to  a  careful  study  of  the  chapter 
in  question  will  be  well  expended. 

The  notation  is  practically  uniform  throughout  the  book, 
and  is  that  used  by  several  standard  authors.  Forces  and  mo- 
ments are  expressed  by  capital  letters,  and  unit  loads  and  stresses 
by  small  letters.  The  coordinate  x  is  measured  along  the  length 

of  a  piece,  the  coordinate  y  in  the  direction  of  variation  of  stress 

iii 


206529 


iv  PREFACE. 

in  a  section,  and  z  is  the  line  of  no  variation  of  stress,  that  is, 
the  line  parallel  to  the  moment  axis. 

One  who  has  mastered  the  subjects  discussed  here  can  use 
the  current  formulas,  the  pocket-book  rules,  and  tables,  not 
blindly,  but  with  discrimination,  and  ought  to  be  prepared  to 
design  intelligently. 


TABLE  OF  CONTENTS. 


PAGE 

INTRODUCTION i 

CHAPTER  I. 
ACTION  OF  A  PIECE  UNDER  DIRECT  FORCE r. 6 

CHAPTER  II. 
MATERIALS 19 

CHAPTER  III. 
BEAMS 38 

CHAPTER  IV. 
MOMENTS  OF  INERTIA  OF  PLANE  AREAS 71 

CHAPTER  \  V. 
TORSION 81 

CHAPTER  VI. 
FLEXURE  AND  DEFLECTION  OF  SIMPLE  BEAMS 87 

CHAPTER  VII. 
RESTRAINED  BEAMS:    CONTINUOUS  BEAMS 107 

CHAPTER  VIII. 
PIECES  UNDER  TENSION 127 

CHAPTER  IX. 
COMPRESSION  PIECES:   COLUMNS,  POSTS,  AND  STRUTS 137 

CHAPTER  X. 
SAFE  WORKING  STRESSES 153 


vi  TABLE   OF  CONTENTS. 

CHAPTER  XI. 

PAGE 

INTERNAL  STRESS:    CHANGE  or  FORM 167 

CHAPTER  XII. 
RIVETS:    PINS 192 

CHAPTER  XIII. 
ENVELOPES:    BOILERS.  PIPES,  DOME 203 

CHAPTER  XIV. 
PLATE  GIRDER 221 

CHAPTER  XV. 
SPRINGS:    PLATES 229 

CHAPTER  XVI. 
REINFORCED  CONCRETE 235 


NOTATION. 

by   breadth  of  rectangular  beam. 
C,  shearing  modulus  of  elasticity. 
d,   diameter. 

£,  modulus  of  elasticity,  Young's  modulus. 
F,  shear  in  beam. 
/,    unit  stress. 

h,  height  of  rectangular  beam. 
/,   rectangular  moment  of  inertia. 
i,    slope  of  elastic  curve. 
/,  polar  moment  of  inertia. 
k,   a  numerical  coefficient. 
/,    length  of  member. 
^,    unit  change  of  length. 
M,  bending  or  resisting  moment. 
Pj  reaction  of  beam ;  load  on  tie  or  post. 
p,  unit  stress;  unit  pressure  in  envelope,  Ch.  XIII. 
q,    unit  shear. 
R,  radius  of  circle. 

r,    radius  of  gyration;   radius  of  envelope,  Ch.  XIII. 
p,  radius  of  curvature. 
S,  area  of  cross-section. 

T,  torsional  moment;  stress  in  envelope,  Ch.  XIII. 
v,   deflection  of  beam. 
W,  concentrated  load  on  beam. 
w,   intensity  of  distributed  load  on  beam. 
y\j  distance  from  neutral  axis  to  extreme  fibre  of  beam. 
oc,  y,  z,  coordinates  of  length,  depth,  and  breadth  of  beam. 

vii 


STRUCTURAL  MECHANICS. 


INTRODUCTION. 

i.  External  Forces. — The  engineer,  in  designing  a  new 
structure,  or  critically  examining  one  already  built,  determines 
from  the  conditions  of  the  case  the  actual  or  probable  external 
forces  which  the  structure  is  called  upon  to  resist.  He  may 
then  prepare,  either  by  mathematical  calculations  or  by  graphical 
methods,  a  sheet  which  shows  the  maximum  and  minimum 
direct  forces  of  tension  and  compression  which  the  several  pieces 
or  parts  of  the  structure  are  liable  to  experience,  as  well  as  the 
bending  moments  on  such  parts  as  are  subjected  to  them. 

These  forces  and  moments  are  determined  from  the  require- 
ments of  equilibrium,  if  the  pieces  are  at  rest.  For  forces  acting 
in  one  plane,  a  condition  which  suffices  for  the  analysis  of  most 
cases,  it  is  necessary  that,  for  the  structure  as  a  whole,  as  well 
as  for  each  piece,  there  shall  be  no  tendency  to  move  up  or  down, 
to  move  to  the  right  or  left,  or  to  rotate.  These  limitations 
are  usually  expressed  in  Mechanics  as,  that  the  sum  of  the  X 
forces,  the  sum  of  the  Y  forces,  and  the  sum  of  the  moments 
shall  each  equal  zero. 

If  the  structure  is  a  machine,  the  forces  and  moments  in 
action  at  any  time,  and  their  respective  magnitudes,  call  for  a 
consideration  of  the  question  of  acceleration  or  retardation  of 
the  several  parts  and  the  additional  maximum  forces  and 
moments  called  into  action  by  the  greatest  rate  of  change  of 


2  STRUCTURAL  MECHANICS. 

motion  at  any  instant.  Hence  the  weight  or  mass  of  the  mov- 
ing part  or  parts  is  necessarily  taken  into  account. 

Finally,  noting  the  rapidity  and  frequency  of  the  change 
of  force  and  moment  at  any  section  of  any  piece  or  connection, 
the  engineer  selects,  as  judgment  dictates,  the  allowable  stresses 
of  the  several  kinds  per  square  inch,  making  allowance  for  the 
effect  of  impact,  shock,  and  vibration  in  intensifying  their  action, 
and  proceeds  to  find  the  necessary  cross -sections  of  the  parts 
and  the  proportions  of  the  connections  between  them.  As  all 
structures  are  intended  to  endure  the  forces  and  vicissitudes  to 
which  they  are  usually  exposed,  the  allowable  unit  stresses, 
expressed  in  pounds  per  square  inch,  must  be  saje  stresses. 

It  is  largely  with  the  development  of  the  latter  part  of  this 
subject,  after  the  forces  have  been  found  to  which  the  several 
parts  are  liable,  that  this  book  is  concerned. 

2.  Ties,   Struts,   and  Beams. — There  are,   in  general,   three 
kinds  of  pieces  in  a  frame  or  structure :   ties  or  tension  members ; 
columns,  posts,  and  struts  or  compression  members;  and  beams, 
which  support  a  transverse  load  and  are  subject  to  bending  and 
its  accompanying  shear.     A   given  piece  may  also   be,   at   the 
same  time,  a  tie  and  a  beam,  or  a  strut  and  a  beam,  and  at  dif- 
ferent times  a  tie  and  a  strut. 

3.  Relation   of   External   Forces   to  Internal   Stresses. — The 
forces  and  moments  which  a  member  is  called  upon  to  resist, 
and  which  may  properly  be  considered  as  external  to  that  mem- 
ber, give  rise  to  actions  between  all  the  particles  of  material  of 
which  such  a  member  is  composed,  tending  to  move  adjacent 
particles  from,  towards,  or  by  one  another,  and  causing  change 
of  form.     There  result   internal  stresses,   or  resistances   to   dis- 
placement, between  the  several  particles. 

These  internal  stresses,  or  briefly  stresses,  must  be  of  such 
kind,  magnitude,  distribution,  and  direction,  at  any  imaginary 
section  of  a  piece  or  structure,  that  their  resultant  force  and 
moment  will  satisfy  the  requirements  of  equilibrium  or  change 
of  motion  with  the  external  resultant  force  and  moment  at  that 
section;  and  no  stress  per  square  inch  can,  for  a  correct  design, 
be  greater  than  the  material  will  safely  bear.  Hence  may  be 


INTRODUCTION.  3 

determined  the  necessary  area  and  form  of  the  cross-section  at 
the  critical  points,  when  the  resultant  forces  and  moments  are 
known. 

4.  Internal  Stress. — There  are  three  kinds  of  stress,  or  action 
of  adjacent  particles  one  on  the  other,  to  which  the  particles  of 
a  body  may  be  subjected,  when  external  forces  and  its  own  weight 
are  considered,  viz. :  tensile  stress,  tending  to  remove  one  particle 
farther  from  its  neighbor,  and  manifested  by  an  accompanying 
stretch  or  elongation  of  the  body;   compressive  stress,  tending  to 
make  a  particle  approach  its  neighbor,  and  manifested  by  an 
accompanying    shortening    or    compression    of    the    body;     and 
shearing  stress,  tending  to  make  a  particle  move  or  slide  laterally 
with  reference  to  an  adjacent  particle,  and  manifested  by  an 
accompanying  distortion.     Whether  the  stress  produces  change 
of  form,  or  the  attempted  change  of  form  gives  rise  to  internal 
stresses  as  resistances,  is  of  little  consequence;  the  stress  between 
two  particles  and  the  change  of  position  of  the  particles   are 
always  associated,  and  one  being  given  the  other  must  exist. 

5.  Tension  and  Shear,  or  Compression  and  Shear. — If  the 
direction  of  the  stress  is  oblique,  that  is,  not  normal  or  perpen- 
dicular, on  any  section  of  a  body,  the  stress  may  be  resolved  into 
a  tensile  or   compressive   stress  normal  to  that   section,  and  a 
tangential  stress  along  the  section,  which,  from  its  tendency  to 
cause  sliding  of  one  portion  of  the  body  by  or  along  the  section, 
has  been  given  the  name  of  shear,  from  the  resemblance  to  the 
action  of  a  pair  of  shears,  one  blade  passing  by  the  other  along 
the  opposite  sides  of  the  plane  of  section.     Draw  two  oblique 
and  directly  opposed  arrows,  one  on  either  side  of  a  straight  line 
representing   the  trace   of   a   sectional   plane,    decompose  those 
oblique   stresses   normally   and   tangentially   to   the   plane,   and 
notice   the   resulting   directly   opposed   tension   or   compression, 
and  the  shear.     Hence  tension  and  shear,  or  compression  and 
shear,  may  be  found  on  any  given  plane  in  a  body,  but  tension 
and  compression  cannot  simultaneously  occur  at  one  point  in 
a  given  area. 

6.  Sign  of  Stress. — Ties  are  usually  slender  members;    struts 
have  larger  lateral  dimensions.     Longitudinal  tension  tends  to 


4  STRUCTURAL  MECHANICS. 

diminish  the  cross-section  of  the  piece  which  carries  it,  and  hence 
may  conveniently  be  represented  by  — ,  the  negative  sign;  longi- 
tudinal compression  tends  to  increase  the  cross-sectional  area 
and  may  be  called  +  or  positive.  Shear,  being  at  right  angles 
to  the  tension  and  compression  in  the  preceding  illustration,  has 
no  sign;  and  lies,  in  significance,  between  tension  and  com- 
pression. If  a  rectangular  plate  is  pulled  in  the  direction  of  two 
of  its  opposite  sides  and  compressed  in  the  direction  of  its  other 
two  sides,  there  will  be  some  shearing  stress  on  every  plane  of 
section  except  those  parallel  to  the  sides,  and  nothing  but  shear 
on  two  certain  oblique  planes,  as  will  be  seen  later. 

7.  Unit  Stresses. — These  internal  stresses  are  measured  by 
units  of  pounds  and  inches  by  English  and  American  engineers, 
and  are  stated  as  so  many  pounds  of  tension,  compression,  or 
shear  per  square  inch,  called  unit  tension,  compression,  or  shear. 
Thus,  in  a  bar  of  four  square  inches  cross-section,  under  a  total 
pull  of  36,000  pounds  centrally  applied,  the  internal  unit  tension 
is  9,000  pounds  per  square  inch,  provided  the  pull  is  uniformly 
distributed  on  the  particles   adjacent   to   any  cross-section.     If 
the  pull  is  not  central  or  the  stress  not  uniformly   distributed, 
the  average  or  mean  unit  tensile  stress  is  still  9,000  pounds. 

If  an  oblique  section  of  the  same  bar  is  made,  the  total  jorce 
acting  on  the  particles  adjacent  to  the  section  is  the  same  as  before, 
but  the  area  of  section  is  increased;  hence  the  unit  stress,  found 
by  dividing  the  force  by  the  new  area,  is  diminished.  The  stress 
will  also  be  oblique  to  the  section,  as  its  direction  must  be  that 
of  the  force.  When  the  unit  stress  is  not  normal  to  the  plane 
of  section  on  which  it  acts,  it  can  be  decomposed  into  a  normal 
unit  tension  and  a  unit  shear.  See  §  151. 

When  the  stress  varies  in  magnitude  from  point  to  point, 
its  amount  on  any  very  small  area  (the  infinitesimal  area  of  the 
Calculus)  may  be  divided  by  that  area,  and  the  quotient  will 
be  the  unit  stress,  or  the  amount  which  would  exist  on  a  square 
inch,  if  a  square  inch  had  the  same  stress  all  over  it  as  the  very 
small  area  has. 

8.  Unit  Stresses  on  Different  Planes  not  to  be  Treated  as 
Forces. — It  will  be  seen,  upon  inspection  of  the  results  of  analyses 


INTRODUCTION.  5 

which  come  later,  that  unit  stresses  acting  on  different  planes 
must  not  be  compounded  and  resolved  as  if  they  were  forces. 
But  the  entire  stress  upon  a  certain  area,  found  by  multiplying 
the  unit  stress  by  that  area,  is  a  force,  and  this  force  may  be 
compounded  with  other  forces  or  resolved,  and  the  new  force 
may  then  be  divided  by  the  new  area  of  action,  and  a  new  unit 
stress  be  thus  found. 

Some  persons  may  be  assisted  in  understanding  the  analysis 
of  problems  by  representing  in  a  sketch,  or  mentally,  the  unit 
stresses  at  different  parts  of  a  cross-section  by  ordinates  which 
make  up,  in  their  assemblage,  a  volume.  This  volume,  whose 
base  is  the  cross-section,  will  represent  or  be  proportional  to 
the  total  force  on  the  section.  The  position  of  the  resultant 
force  or  forces,  i.e.,  traversing  the  centre  of  gravity  of  the  volume, 
the  direction  and  law  of  distribution  of  the  stress  are  then  quite 
apparent. 


CHAPTER  I. 
ACTION  OF  A  PIECE  UNDER  DIRECT  FORCE. 

9.  Change  of  Length  under  an  Applied  Force. — Let  a  uniform 
bar  of  steel  have  a  moderate  amount  of  tension  applied  to  its 
two  ends.  It  will  be  found,  upon  measurement,  to  have  increased 
in  length  uniformly  throughout  the  measured  distance.  Upon 
release  of  the  tension  the  stretch  disappears,  the  bar  resuming 
its  original  length.  A  second  application  of  the  same  amount  of 
tension  will  cause  the  same  elongation,  and  its  removal  will  be 
followed  by  the  same  contraction  to  the  original  length.  The 
bar  acts  like  a  spring.  This  elastic  elongation  (or  shortening 
under  compression)  is  manifested  by  all  substances  which  have 
definite  form  and  are  used  in  construction;  and  it  is  the  cause 
of  such  changes  of  shape  as  structures,  commonly  considered 
rigid,  experience  under  changing  loads.  The  product  of  the 
elongation  (or  shortening)  into  the  mean  force  that  produced  it 
is  a  measure  of  the  work  done  in  causing  the  change  of  length. 
As  the  energy  of  a  moving  body  can  be  overcome  only  by  work 
done,  the  above  product  becomes  of  practical  interest  in  structures 
where  moving  loads,  shocks,  and  vibrations  play  an  important 
part. 

10.  Modulus  of  Elasticity. — If  the  bar  of  steel  is  stretched 
with  a  greater  force,  but  still  a  moderate  one,  it  is  found  by  care- 
ful measurement  that  the  elongation  has  increased  with  the 
force;  and  the  relationship  may  be  laid  down  that  the  elongation 
per  linear  inch  is  directly  proportional  to  the  unit  stress  on  the 
cross-section  per  square  inch. 

The  ratio  of  the  unit  stress  to  the  elongation  per  unit  of  length 
is  denoted  by  E,  which  is  termed  the  modulus  oj  elasticity  of  the 

6 


ACTION   OF  A   PIECE  UNDER  DIRECT  FORCE.  7 

material,  and  is  based,  in  English  and  American  books,  upon 
the  pound  and  inch  as  units.  If  P  is  the  total  tension  in  pounds 
applied  to  the  cross-section,  S,  measured  in  square  inches,  H 
the  elongation  in  inches,  produced  by  the  tension,  in  the  pre- 
viously measured  length  of  /  inches,  and  /  the  stress  per  square 
inch  of  cross-section, 

E-PJ.  J 

~sri'         ~E- 

Hence,  if  E  has  been  determined  for  a  given  material,  the 
stretch  of  a  given  bar  under  a  given  unit  stress  is  easily  found. 

Since  the  elongation  per  unit  of  length,  A,  is  merely  a  ratio 
and  is  the  same  whatever  system  of  units  is  employed,  E  will  be 
expressed  in  the  same  units  as  /. 

Example. — A  bar  of  6  sq.  in.  section  stretches  0.085  m-  'm  a  measured 
length  of  1 20  in.  under  a  pull  of  120,000  Ib. 

120,000X120 

E=  —         — - - — -=28,200,000  Ib.  per  sq.  m. 
6X0.085 

If  the  stress  were  compressive,  a  similar  modulus  would 
result,  which  will  be  shown  presently  to  agree  with  the  one  just 
derived. 

If  one  particle  is  displaced  laterally  with  regard  to  its  neighbor, 
under  the  action  of  a  shearing  stress,  a  modulus  of  shearing  elas- 
ticity will  be  obtained,  denoted  by  C,  the  ratio  of  the  unit  shear 
to  the  angle  of  distortion.  See  §  173. 

ii.  Stress-stretch  Diagram. — The  elongations  caused  in  a 
certain  bar,  or  the  stretch  per  unit  of  length,  may  be  plotted 
as  abscissas,  and  the  corresponding  forces  producing  the  stretch, 
or  the  unit  stresses  per  square  inch,  may  be  used  as  ordinates, 
denning  a  certain  curve,  as  represented  in  Fig.  i.  This  curve 
can  be  drawn  on  paper  by  the  specimen  itself,  when  in  the  testing- 
machine,  if  the  paper  is  moved  in  one  direction  to  correspond 
with  the  movement  of  the  poise  on  the  weighing  arm,  and  the 
pencil  is  moved  at  right  angles  by  the  stretch  of  the  specimen. 

A  similar  diagram  can  be  made  for  a  compression  specimen, 
and  may  be  drawn,  in  the  diagonally  opposite  quadrant.  Pull 


8  STRUCTURAL  MECHANICS. 

will  then  be  rightly  represented  as  of  opposite  sign  to  thrust, 
and  extension  will  be  laid  off  in  the  opposite  direction  to  shorten- 
ing or  compression. 

12.  Work     of     Elongation. — If    the    different    unit    stresses 
applied  to  the  bar  are  laid  off  on  O  Y  as  ordinates  and  the  result- 
ing stretches  per  unit  of  length  on  O  X  as  abscissas,  the  portion 
of  the  diagram  near  the  origin  will  be  found  to  be  a  straight  line, 
more  or  less  oblique,  according  to  the  scale  by  which  the  elonga- 
tions are  platted.     The  elongation  varies  directly  as  the  unit 
stress,  beginning  with  zero.     Hence  the  mean  force  is  JP,  and 
the  work  done  in  stretching  a  given  bar  with  a  given  force,  if 
the  limit  of  elastic  stretch  is  not  exceeded,  is 

P          P2l 

VV  OlT-K.  *  A*'  TT~r*~* 

2  2ES 

It  may  be  seen  that  the  work  done  in  stretching  the  bar  is 
represented  by  the  area  included  between  the  base  line  or  axis, 
the  curve  O  A,  and  the  ordinate  at  A.  It  also  appears  that  E 
may  be  looked  upon  as  the  tangent  of  the  angle  X  O  A.  A 
material  of  greater  resistance  to  elongation  will  give  an  angle 
greater  than  X  O  A  and  vice  versa. 

Example. — A  bar  20  ft.  =  240  in.  long  and  3  sq.  in.  in  section  is 
to  have  a  stress  applied  of  10,000  Ib.  per  sq.  in.;  if  £=28,000,000, 
the  work  done  on  the  bar  will  be 

30,000-  30,000-240 
—     — —  -  =1,286  m.-lb., 

2-28,000,000-3 

and  the  stretch  will  be  1,286^15,000=0.086  in. 

13.  Permanent  Set.— While   the   unit   stress   may   be   grad- 
ually   increased    with    corresponding    increase    of    stretch,    and 
apparently  complete  recovery  of  original  length  when  the  bar 
is  released,  there  comes  a  time  when  very  minute  and  delicate 
measurements  show  that  the  elongation  has  increased  in  a  slightly 
greater  degree  than  has  the  stress.     The  line  O  A  at  and  beyond 
such  a  point  must  therefore  be  a  curve,  concave  to  the  axis  of  X. 
If  the  piece  is  now  relieved  from  stress,  it  will  be  found  that  the 


ACTION  OF  A    PIECE   UNDER  DIRECT  FORCE.  9 

bar  has  become  permanently  lengthened.  The  amount  of  this 
increase  of  length  after  removal  of  stress  is  called  set,  or  perma- 
nent set,  and  the  unit  stress  for  which  a  permanent  set  can  first 
be  detected  is  known  as  the  elastic  limit.  As  the  elongation 
itself  is  an  exceedingly  small  quantity,  even  when  measured  in 
a  length  of  many  inches,  and  the  permanent  set  is,  in  the  begin- 
ning, a  quantity  far  smaller  and  hence  more  difficult  of  deter- 
mination, the  place  where  the  straight  line  OA  first  begins  to 


.38      X 


curve  is  naturally  hard  to  locate,  and  the  accurate  elastic  limit 
is  therefore  uncertain.  Some  contend  that  O  A  itself  is  a  curve 
of  extreme  flatness.  The  common  or  commercial  elastic  limit  lies 
much  farther  up  the  curve,  where  the  permanent  set  becomes 
decidedly  notable. 

If,  after  a  certain  amount  of  permanent  set  has  occurred 
in  a  bar,  and  the  force  which  caused  it  has  been  removed,  a 
somewhat  smaller  force  is  repeatedly  applied  to  the  bar,  the 
piece  will  elongate  and  contract  elastically  to  the  new  length, 


10  STRUCTURAL  MECHANICS. 

i.e.j  old  length  plus  permanent  set,  just  as  if  the  unit  stress  were 
below  the  elastic  limit. 

14.  Yield-point. — The  unit    stress  increasing,  the  elongation 
increases  and  the  permanent  set  increases  until  a  unit  stress  B 
is  reached,  known  as  the  yield-point  (or  commercial  elastic  limit, 
or  common  elastic  limit),  which  causes  the  bar  to  yield  or  draw 
out  without  increase  of  force,  and,  as  the  section  must  decrease, 
apparently    with    decreasing    power   of   resistance.     There    will 
then  be  a  break  of  continuity  in  the  graphic  curve.     A  decided 
permanent  elongation  of  the  bar  takes  place  at  this  time — suf- 
ficient to  dislodge  the  scale  from  the  surface  of  a  steel  bar,  if 
left  as  it  comes  from  the  rolls  or  hammer.     The  weighing  beam 
of    the    testing-machine    falls,    from    the    diminished    resistance 
just  referred  to,  and  remains  stationary  while  the  bar  is  elongating 
for  a  sensible  interval  of  time.     Hence,  for  steel,  the  yield-point, 
or  common  elastic  limit,  is  easily  determined  by  what  is  known 
as  the  "drop  of  the  beam."     The  remainder  of  the  curve,  up  to 
the  breaking-point,  is  shown  in  the  figure. 

15.  Elastic    Limit   Raised. — For    stresses    above    the    yield- 
point  also,  a  second  application  and  release  of  stress  will  give 
an  elastic  elongation  and  contraction  as  before  the  occurrence 
of  set,  as  shown  by  line  E  F,  so  that  a  new  elastic  limit  may  be 
said  to  be  established.     The  stretch  due  to  any  given  stress  may 
be  considered  to  be  the  elastic  elongation  plus  the  permanent 
set;   and,  for  repetitions  of  lesser  forces,  the  bar  will  give  a  line 
parallel  to  O  A,  as  if  drawn  from  a  new  origin  on  O  X,  distant 
from  O  the  amount  of  the  permanent  set. 

If  the  line  O  A  is  prolonged  upwards,  it  will  divide  each 
abscissa  into  two  parts,  of  which  that  on  the  left  of  O  A  will 
be  the  elastic  stretch,  and  that  on  the  right  of  O  A  the  permanent 
set  for  a  given  unit  stress. 

16.  Work  of  Elongation,  for  Stress  above  Yield-point. — The 
area  below  the  curve,  and  limited  by  any  ordinate    G  D,  will 
be  the  work  done  in  stretching  the  bar  with  a  force  represented 
by  the  product  of  that  ordinate  into  the  bar's  cross-section,  and 
if  a  line  be  drawn  from  the  upper  end  of  that  ordinate  parallel 
to  O  A,  the  triangle  C  D  G  will  give  the  work  done  in  elastic 


ACTION  OF  A   PIECE   UNDER  DIRECT  FORCE.  II 

stretch  and  the  quasi -parallelogram  O  B  D  C  will  show  the 
permanent  work  of  deformation  done  on  the  bar.  It  should 
be  remembered  that,  as  the  bar  stretches,  the  section  decreases, 
and  that  the  unit  stress  cannot  therefore  be  strictly  represented 
by  P-^-S,  if  S  is  the  original  cross-section.  The  error  is  not  of 
practical  consequence  for  this  discussion. 

17.  Ultimate  or  Breaking  Strength. — If  the  force  applied  in 
tension  to  the  bar  is  increased,  a  point  will  next  be  reached  where 
a  repeated  application  of  the  same  force  causes  a  successive  increase 
in  the  permanent  elongation.  As  this  phenomenon  means  a 
gradual  drawing  out,  final  failure  by  pulling  asunder  is  only 
a  matter  of  a  greater  or  less  number  of  applications  of  the  force. 
While  the  bar  is  apparently  breaking  under  this  force,  the  rapid 
diminution  of  cross-section  near  the  breaking  point  actually  gives 
a  constantly  rising  unit  stress,  as  is  seen  by  the  dotted  curve  of 
the  figure. 

If,  however,  the  force  is  increased  without  pause  from  the 
beginning,  the  breaking  force  will  be  higher,  as  might  be  ex- 
pected; since  much  work 'of  deformation  is  done  upon  the  bar 
before  fracture.  The  bar  would  have  broken  under  a  somewhat 
smaller  force,  applied  statically  for  a  considerable  time. 

The  elongation  of  the  bar  was  uniform  per  unit  of  length 
during  the  earlier  part  of  the  test.  There  comes  a  time  when  a 
portion  or  section  of  the  bar,  from  some  local  cause,  begins  to 
yield  more  rapidly  than  the  rest.  At  once  the  unit  stress  at 
that  section  becomes  greater  than  in  the  rest  of  the  bar,  by  reason 
of  decrease  of  cross-section,  and  the  drawing  out  becomes  in- 
tensified, with  the  result  of  a  great  local  elongation  and  necking 
of  the  specimen  and  an  assured  final  fracture  at  that  place.  If 
the  bar  were  perfectly  homogeneous,  and  the  stress  uniformly 
distributed,  the  bar  ought  to  break  at  the  middle  of  the  length, 
where  the  flow  of  the  metal  is  most  free. 

It  is  customary  to  determine,  and  to  require  by  specification, 
in  addition  to  elastic  limit  and  ultimate  strength  (on  one  con- 
tinuous application  of  increasing  load),  the  per  cent,  of  elonga- 
tion after  fracture  (which  is  strictly  the  permanent  set)  in  a  cer- 
tain original  measured  length,  usually  eight  inches,  and  the  per 


12  STRUCTURAL   MECHANICS. 

cent,  of  reduction  of  the  original  area,  after  fracture,  at  the  point 
of  fracture.  As  the  measured  length  must  include  the  much 
contracted  neck,  the  average  per  cent,  of  elongation  is  given 
under  these  conditions.  A  few  inches  excluding  the  neck  would 
show  less  extension,  and  an  inch  or  two  at  the  neck  would  give 
a  far  higher  per  cent,  of  elongation.  The  area  between  the  axis 
of  X,  the  extreme  ordinate,  and  the  curve  will  be  the  work  of 
fracture,  if  S  is  considered  constant,  and  will  be  a  measure  of 
the  ability  of  the  material  to  resist  shocks,  blows,  and  vibrations 
before  fracture.  It  is  indicative  of  the  toughness  or  ductility  of 
the  material. 

The  actual  curve  described  by  the  autographic  attachment 
to  a  testing-machine  is  represented  by  the  full  line;  the  real 
relation  of  stress  per  square  inch  to  the  elongation  produced, 
when  account  is  taken  of  the  progressive  reduction  of  sectional 
area,  is  shown  by  the  dotted  line.  The  yield-point,  or  common 
elastic  limit,  is  very  marked,  there  appearing  to  be  a  decided 
giving  way  or  rearrangement  of  the  particles  at  that  value  of 
stress.  The  true  elastic  limit  is  much  below  that  point. 

18.  Effect  of  a  Varying  Cross-section. — If  a   test   specimen 
is  reduced  to  a  smaller  cross-section,  by  cutting  out  a  curved 
surface,  for  only  a  short  distance  as  compared  with  its  transverse 
dimensions,  it  will  show  a  greater  unit  breaking  stress,  as  the 
metal  does  not  flow  freely,  and  lateral  contraction  of  area  is 
hindered.     But,  if  the  portion  of  reduced  cross-section  joins  the 
rest  of  the  bar  by  a  shoulder,  the  apparent  strength  is  reduced, 
owing  to  a  concentration  of  stress  on  the  particles  at  the  corner 
as  the  unit  stress  suddenly  changes  from  the  smaller  value  on  the 
larger  section  to  the  greater  unit    stress  on  the   smaller  cross- 
section. 

19.  Compression  Curve. — A  piece  subjected  to  compression 
will  shorten,  the  particles  being  forced  nearer  together,  and  the 
cross-section  will  increase.     It  might  be  expected,  and  is  found 
by  experiment  to  be  the  case,  that,  in  the  beginning,  the  resistance 
of  the  particles  to  approach  would  be  like  their  resistance  to 
separation  under  tension,  so  that  the  tension  diagram  might  be 
prolonged  through  the  origin  into  the  third  quadrant,  reversing 


ACTION  OF   A   PIECE   UNDER  DIRECT  FORCE.  13 

the  sign  of  the  ordinate  which  represents  unit  stress  and  of  the 
abscissa  which  shows  the  corresponding  change  of  length.  As 
this  part  of  the  diagram  is  a  straight  line,  it  follows  that  the  value 
of  E,  the  elastic  modulus  for  compression,  is  the  same  as  that 
for  tension.  After  passing  the  yield-point  the  phenomena  of  com- 
pression are  not  so  readily  determined,  as  fracture  or  failure  by 
compressive  stress  is  not  a  simple  matter,  and  the  increase  of 
sectional  area  in  a  chort  column  of  ductile  material  will  interfere 
with  the  experiment.  In  long  columns  and  with  materials  not 
ductile,  failure  takes  place  in  other  ways,  as  will  be  explained 
later. 

The  compression  curve  is  here  shown  in  the  same  quadrant 
with  the  tension  curve  for  convenience  and  comparison. 

20.  Resilience.  —  By  definition,  §  10,  if  /  is  the  unit  stress 
per  square  inch  and  XI  the  stretch  of  a  bar  of  length  /,  in  inches, 
the  modulus  of  elasticity  E  =  /-J-^,  provided  /  does  not  exceed 
the  elastic  limit.  Also  the  work  done  in  stretching  a  bar  inside 
the  elastic  limit,  by  a  force  P,  gradually  applied,  that  is,  beginning 
with  zero  and  increasing  with  the  stretch,  is  the  product  of  the 
mean  force,  JP,  into  the  stretch,  or 


Work 

2     E       2     E 

The  amount  of  work  which  must  be  done  upon  a  piece  in  order 
to  produce  the  safe  unit  stress,  /,  in  it  is  the  resilience  of  the  piece. 
SI  is  the  volume  of  the  bar;  /2-4-E  is  called  the  modulus  of  re- 
silience, when  /  is  the  elastic  limit,  or  sometimes  the  maximum 
safe  unit  stress.  This  modulus  depends  upon  the  quality  of  the 
material,  and,  as  it  is  directly  proportional  to  the  amount  of 
work  that  can  safely  be  done  upon  the  bar  by  a  load,  it  is  a 
measure  of  the  capacity  of  a  certain  material  for  resisting  or 
absorbing  shock  and  impact  without  damage.  For  a  particular 
piece,  the  volume  SI  is  also  a  factor  as  above.  A  light  structure 
will  suffer  more  from  sudden  or  rapidv  loading  than  will  a 
heavier  one  of  the  same  material,  if  proportioned  for  the  same 
unit  stress. 


14  STRUCTURAL  MECHANICS. 

21.  Work  Done  Beyond  the  Elastic  Limit. — The  work  done 
in  stretching  a  bar  to  any  extent  is,  in   Fig.  i,  the  area  in  the 
diagram  between  the  curve  from  the  origin  up  to  any  point, 
the  ordinate  to  that  point,  and  the  axis  of  abscissas,  provided 
the  ordinate  represents  P,  and  the  abscissa  the  total  stretch. 

Further,  it  may  be  seen  from  the  figure  that,  if  a  load  applied 
to  the  bar  has  exceeded  the  yield-point,  the  bar,  in  afterwards 
contracting,  follows  the  line  F  E;  and,  upon  a  second  applica- 
tion of  the  load,  the  right  triangle  of  which  this  line  is  the  hypothe- 
nuse  will  be  the  work  done  in  the  second  application,  a  smaller 
quantity  than  for  the  first  application.  But,  if  the  load,  in  its 
second  and  subsequent  applications,  possesses  a  certain  amount 
of  energy,  by  reason  of  not  being  gently  or  slowly  applied,  this 
energy  may  exceed  the  area  of  the  triangle  last  referred  to,  with 
the  result  that  the  stress  on  the  particles  of  the  bar  may  become 
greater  than  on  the  first  application.  Indeed  it  is  conceivable 
that  this  load  may  be  applied  in  such  a  way  that  the  resulting 
unit  stress  may  mount  higher  and  higher  with  repeated  applica- 
tions of  load,  until  the  bar  is  broken  with  an  apparent  unit  stress 
P+S,  far  less  than  the  ultimate  strength,  and  one  which  at  first 
was  not  much  above  the  yield-point.  If  the  load  in  its  first 
application  is  above  the  yield-point  of  the  material,  and  it  is 
repeated  continuously,  rupture  will  finally  occur. 

What  is  true  for  tensile  stresses  is  equally  true  for  compressive 
stresses,  except  that  the  ultimate  strength  of  ductile  materials 
under  compression  is  uncertain  and  rather  indefinite. 

22.  Sudden  Application  of  Load. — If  a  steel  rod,  10  feet  = 
1 20   inches   long,    and   one   square    inch    in  section,    with    E  = 
28,000,000,  is  subjected  to  a  force  increasing    gradually  from   o 
to  12,000  Ib.    longitudinal   tension,  its   stretch   will  be    1 2,000  X 
120-^28,000,000  =  0.05   in.,  and  the  work  done  in  stretching  the 
bar  will  be  1X12,000X0.05=300  in.-lb. 

But  if  the  12,000  Ib.  is  suddenly  applied,  as  by  the  extremely 
rapid  loading  of  a  structure  of  which  the  rod  forms  a  part,  or 
by  the  quick  removal  of  a  support  which  held  this  weight  at  the 
lower  end  of  the  rod,  the  energy  due  to  a  fall  of  0.05  in.  is  12,000  X 
0.05  =  600  in.-lb.,  while  the  work  done  upon  the  rod  is  but  300  in.-lb. 


ACTION  OF  A    PIECE   UNDER  DIRECT  FORCE.  15 

as  before.  The  excess  of  300  in.-lb.  of  energy  must  be  absorbed 
by  the  rod  and  it  will  continue  to  stretch  until  the  energy  due  to 
the  fall  equals  the  work  done  upon  the  rod  or  until  it  has  stretched 
o.io  in.  The  stress  in  the  rod  is  then  24,000  Ib.  or  twice  the 
suddenly  applied  load;  the  energy  due  to  the  fall  is  12,000X0.10 
and  the  work  done  upon  the  rod  is  JX  24,000X0.10  =  1,200  in.-lb. 
As  equilibrium  does  not  exist  between  the  external  force  and  the 
internal  stress,  the  rod  will  contract  and  then  undergo  a  series  of 
longitudinal  vibrations  of  decreasing  amplitudes,  finally  settling 
down  to  a  stretch  of  0.05  in.,  when  the  extra  work  of  accelera- 
tion has  been  absorbed.  The  work  of  acceleration  on  the  mass 
of  the  bar  is  neglected. 

A  load  applied  to  a  piece  with  absolute  suddenness  produces 
twice  the  deformation  and  twice  the  stress  which  the  same  load 
does  if  applied  gradually.  Stresses  produced  by  moving  loads 
on  a  structure  are  intermediate  in  effect  between  these  two  extremes^ 
depending  upon  rapidity  or  suddenness  of  loading.  Hence  it  is 
seen  why  the  practice  has  arisen  of  limiting  stresses  due  to  moving 
loads  apparently  to  only  one-half  of  the  values  permitted  for 
those  caused  by  static  loading. 

For  resilience  or  work  done  in  deflection  of  beams,  see  §  100. 

23.  Granular  Substances  under  Compression. — Failure  by 
Shearing  on  Oblique  Planes.  Blocks  of  material,  such  as  cast 
iron,  sandstone,  or  concrete,  when  subjected  to  compression, 
frequently  give  way  by  fracturing  on  one  or  more  oblique  planes 
which  cut  the  block  into  two  wedges,  or  into  pyramids  and  wedges. 
The  pyramids  may  overlap,  and  their  bases  are  in  the  upper 
and  lower  faces  of  the  block.  This  mode  of  fracture,  peculiar 
to  granular  substances,  of  comparatively  low  shearing  resistance, 
can  be  discussed  as  follows: 

If  a  short  column,  Fig.  2,  of  cross-section  S  is  loaded  cen- 
trally with  P,  the  unit  compression  on  the  right  section  will  be 
pi=P+S,  and  if  the  short  column  gives  way  under  this  load, 
this  value  of  pi  is  commonly  considered  the  crushing  strength 
of  the  material.  While  it  doubtless  is  the  available  crushing 
strength  of  this  specimen,  it  may  by  no  means  represent  the  maxi- 
mum resistance  to  crushing  under  other  conditions. 


i6 


STRUCTURAL   MECHANICS. 


If  pi=P+S  is  the  unit  thrust  on  the  right  section,  it  is  seen, 
from  §  151,  that,  on  a  plane  making  an  angle  6  with  the  right 
section,  the  normal  unit  stress  pn  =  pi  cos2  0,  and  the  tangential 
unit  stress  q=pi  sin  0  cos  #.  If  m= coefficient  of  frictional  re- 
sistance of  the  material  to  sliding,  the  resistance  per  square  inch 


Fig.  2 


to  sliding  along  this  oblique  plane  will  be  mpn  =  mpi  cos2  6,  and 
the  portion  of  the  unit  shearing  stress  tending  to  produce  fracture 
along  this  plane  will  be  q  —  mpn  =  pi  (sin  6  cos  6  —  m  cos2  0)  . 

Fracture  by  shearing,  if  it  occurs,  will  take  place  along  that 
plane  for  which  the  above  expression  is  a  maximum,  or  d(q  —  mpn) 
+d6=o.  Differentiating  relatively  to  #, 

pi  (cos2  6  —  sin2  6  +  2  m  sin  6  cos  6)  =  o  ; 
sin2  6  —  2m  sin  6  cos  0  +  m2  cos2  6  =  cos2  0  +  m2  cos2  0; 
sin  6  —  m  cos  6  =  cos  #\/(i 


+  m2). 


If  m  were  zero,  0  max.  would  be  45°.  Therefore  the  plane  of 
fracture  always  makes  an  angle  greater  than  45°  with  the  right 
section.  As  6  may  be  negative  as  well  as  positive,  fracture  tends 
to  form  pyramids  or  cones. 

Example.  —  A  rectangular  prism  of  cast  iron,  2  in.  high  and  square 
section=i.o5  sq.  in.,  sheared  off  under  a  load  of  97,000  lb.,  or  92,380 


ACTION  OF  A    PIECE   UNDER  DIRECT  FORCE.  l^ 

Ib.  compression  per  sq.  in.  of  cross-section,  at  an  angle  whose  tangent 
was  1.5,  or  56°  19'. 


;  2.25  —  3Ww=i     w;  7^=0.42. 

Sin  #=0.8321,    cos  ^=0.5546,  sin  6  cos  #=0.461,  cos2  #=0.308. 
92,380(0.461  —  0.42  X  0.308)  =  30,680  Ib. 

The  coefficient  of  friction  is  0.42,  and  the  shear  30,680  Ib.  per  sq.  in. 
The  crushing  strength  of  a  short  block  would  have  exceeded  consid- 
erably the  above  92,400  Ib. 

Since  this  deviation  of  the  plane  of  fracture  from  45°  is  due 
to  a  resistance  analogous  to  friction,  it  follows  that,  when  a 
column  of  granular  material,  and  of  moderate  length,  gives  way 
by  shearing,  the  value  pi  will  be  only  that  compressive  stress 
which  is  compatible  with  the  unit  shearing  strength,  while  its 
real  compressive  strength  in  large  blocks  will  be  much  higher. 

The  same  phenomenon  is  exhibited  by  blocks  of  sandstone 
and  of  concrete.  Tests  of  cubes  and  flat  pieces  yield  higher 
results  than  do  those  of  prisms  of  the  same  cross-section  and 
having  a  considerably  greater  height. 

24.  Ductile    Substances   under  Compression.  —  Wrought  iron, 
and  soft  and  medium  steel,  as  well  as  other  ductile  substances, 
tested  in  short  blocks  in  compression,  bulge  or  swell  in  trans- 
verse   dimensions,    and    do    not    fracture.     Hence    the    ultimate 
compressive  strength  is  indefinite. 

25.  Fibrous    Substances    under    Compression.  —  Wood    and 
fibrous  substances  which  have  but  small  lateral  cohesion  of  the 
fibres,  when  compressed  in  short  pieces  in  the  direction  of  the 
same,  separate  into  component  fibres  at  some  irregular  section, 
and  the  several  fibres  fail  laterally  and  crush. 

26.  Vitreous  Substances  under   Compression.  —  Vitreous  sub- 
stances, like  glass  and  vitrified  bricks,  tend  to  split  in  the  direc- 
tion of  the  applied  force. 

27.  Resistance  of  Large  Blocks.  —  The  resistance  per  square 
inch  of  a  cube  to  compression  will  depend  upon  the  size  of  the 
cube.     As   the    unit    stress    and   trm  resulting   deformation   are 
associated,  as  noted  in  §  4,  it  follows  that  the  unit  compressive 
stress  will  be  greatest  at  the  centre  of  the  compressed  surface 


1 8  STRUCTURAL  MECHANICS. 

and  least  at  the  free  edges  where  lateral  movement  of  the  particles 
is  less  restrained.  Hence,  the  larger  the  cube,  the  greater  the 
mean  or  apparent  strength  per  square  inch.  Large  blocks  of 
stone,  therefore,  have  a  greater  average  sustaining  power  per 
square  inch  than  is  indicated  by  small  test  specimens,  other  things 
being  equal. 

The  same  inference  can  be  drawn  as  to  resistance  of  short 
pieces  to  tension  as  compared  with  longer  pieces  of  the  same 
cross-section. 

A  uniform  compression  over  any  cross-section  of  a  large  post 
or  masonry  pier,  when  the  load  is  centrally  applied  to  but  a  small 
portion  of  the  top  can  be  realized  only  approximately ;  the  same 
thing  is  probably  true  of  the  foundation  below  the  pier.  The 
resisting  capacity  of  the  material,  if  earth,  is  thereby  enhanced; 
for  the  tendency  to  escape  laterally  at  the  edges  of  the  foundation 
is  not  so  great  as  would  be  the  case  if  the  load  were  equally  severe 
over  the  whole  base. 

Beveling  the  edges  of  the  compressed  face  of  a  block  will 
increase  the  apparent  resistance  of  the  material  by  taking  the 
load  from  the  part  least  able  to  stand  the  pressure.  The  un- 
loaded perimeter  may  then  act  like  a  hoop  to  the  remainder. 

Examples. — i.  A  round  bar,  i  in.  in  diameter  and  10  ft.  long, 
stretches  0.06  in.,  under  a  pull  of  10,000  Ib.  What  is  the  value  of  E? 
What  is  the  work  done  ?  25,464,733;  300  in.-lb. 

2.  If  the  elastic  limit  of  the  bar  is  reached  by  a  tension  of  30,000 
Ib.  per  sq.  in.,  what  is  the  work  done  or  the  resilience  of  the  bar? 

1,666  in.-lb. 

3.  An  iron  rod,  £=29,000,000,  hangs  in  a  shaft  1,500  ft.  deep. 
What  will  be  the  stretch?  1.55  in. 

4.  A  certain  rod,  22  ft.  long,  and  having  £=28,000,000,  is  to  be 
adjusted  by  a  nut  of  8  threads  to  the  inch  to  an  initial  tension  of  10,000 
Ib.  per  sq.  in.     If  the  connections  were  rigid,  how  much  of  a  revolution 
ought  to  be  given  to  the  nut  after  it  fairly  bears  ?  °-75- 

5.  Can  a  weight  of  20,000  Ib.  be  lifted  by  cooling  a  steel  bar,  i  in. 
sq.,  from  2i2p  to  62°  F.  ?     Coefficient  of  expansion =0.0012  for  180°; 
E=  29,000,000. 

6.  A  steel  eye-bar,  80  in.  long  and  2  in.  scf'.,  fits  on  a  pin  at  each 
end  with  -fa  in.  play.     What  will  be  the    tension  in  the    bar,  if  the 
temperature  falls  75°  F.  and  the  pins  do  not  yield? 

7,250  Ib.  per  sq.  in. 

7.  A  cross-grained  stick  of  pine,  i  sq.  in.  in  section,  sheared  off  at 
an  angle  of  about  66P  with  the  right  section  under  a  compressive  load 
of  3,200  Ib.     If  the  coefficient  of  friction  is  0.5,  what  is  the  unit  shearing 
stress  of  the  section,  the  actual  irregular  area  being  2.9  sq.  in.  ? 

780  Ib. 


CHAPTER  II. 
MATERIALS. 

28.  Growth  of  Trees. — Trees  from  which  lumber  is  cut  grow 
by  the  formation  of  woody  fibre  between  the  trunk  and  the  bark, 
and  each  annual  addition  is  more  or  less  distinctly  visible  as  a 
ring.  Each  ring  is  made  up  of  a  light,  porous  part,  the  spring 
wood,  and  a  darker,  dense  part,  the  summer  wood.  Since  the 
latter  is  firm  and  heavy  it  determines  to  a  large  extent  the  weight 
and  strength  of  the  timber.  The  sap  circulates  through  the  newer 
wood,  and  in  most  trees  the  heart- wood,  as  it  is  called,  can  be 
easily  distinguished  from  the  sap-wood.  The  former  is  con- 
sidered more  strong  and  durable,  unless  the  tree  has  passed  its 
prime.  The  heart  then  deteriorates.  Sap-wood,  in  timber  ex- 
posed to  the  weather,  is  the  first  to  decay. 

Branches  increase  in  size  by  the  addition  of  rings,  as  does 
the  trunk;  hence  a  knot  is  formed  at  the  junction  of  the  branch 
with  the  trunk.  The  knot  begins  where  the  original  bud  started, 
and  increases  in  diameter  towards  the  exterior  of  the  trunk,  as 
the  branch  grows.  The  grain  of  the  annual  growth,  formed 
around  the  junction  of  the  branch  with  the  trunk,  is  much  dis- 
torted. Hence  timber  that  contains  large  knots  is  very  much 
weaker  than  straight-grained  timber.  Even  small  knots  deter- 
mine the  point  of  fracture  when  timber  is  experimentally  tested 
for  strength.  When  a  branch  happens  to  die,  but  the  stub  re- 
mains, and  annual  rings  are  added  to  the  trunk,  a  dead  or  loose 
knot  occurs  in  the  sawed  timber;  such  a  knot  is  considered  a 
defect,  as  likely  to  let  in  moisture  and  start  decay. 

As  forest  trees  grow  close  together,  the  branches  die  succes- 
sively from  below  from  lack  of  sunlight;  such  trees  develop 

19 


20  STRUCTURAL  MECHANICS. 

straight  trunks  of  but  little  taper,  free  from  any  knots,  except 
insignificant  ones  immediately  around  the  centre,  and  yield 
straight-grained,  clear  lumber.  A  few  trees,  like  hemlock,  some- 
times have  their  fibres  running  in  a  spiral,  and  hence  yield  cross- 
grained  timber.  Trees  that  grow  in  open  spaces  have  large  side 
limbs,  and  the  lumber  cut  from  them  has  large  knots. 

29.  Shrinkage  of  Timber. — If  a  log   is  stripped  of  its  bark 
and  allowed  to  dry  or  season,  it  will  be  found  that  the  contrac- 
tion or  shrinkage  in  the  direction  of  .the  radius  is  practically 
nothing.     There  are  numerous  bundles  or  ribbons  of  hard  tissue 
running  radially  through  the  annual  rings  which  appear  to  pre- 
vent such  shrinkage.     Radial  cracks,  running  in  to  a  greater  or 
less  distance,  indicate  that  the  several  rings  have  yielded  to  the 
tension  set  up  by  the  tendency  to  shrink  circumferentially.     Sawed 
timber  of  any  size  is  likely  to  exhibit  these  season  cracks.     Such 
cracks  are  blemishes  and  may  weaken  the  timber  when  used 
for  columns  or  beams.     By  slow  drying,  and  by  boring  a  hole 
through  the  axis  to  promote  drying  within,  the  tendency  to  form 
season  cracks  may  be  diminished. 

A  board  sawed  radially  from  a  log  will  not  shrink  in  width, 
and  will  resist  wear  in  a  floor.  Such  lumber  is  known  as  quarter- 
sawed.  A  board  taken  off  near  the  slab  will  shrink  much  and 
will  tend  to  warp  or  become  concave  on  that  side  which  faced 
the  exterior  of  the  log.  For  that  reason,  and  because  the  annual 
rings  have  less  adhesion  than  the  individual  fibres  have,  all  boards 
exposed  to  wear,  as  in  floors,  should  be  laid  heart-side  down. 

30.  Decay  of  Wood. — Timber  exposed  to  the  weather  should 
be  so  framed  together,  if  possible,  that  water  will  not  collect  in 
joints  and  mortises,  and  that  air  may  have  ready  access  to  all 
parts,  to  promote  rapid  drying  after  rain.    The  end  of  the  grain 
should  not  be  exposed  to  the  direct  entrance  of  water,  but  should 
be  covered,  or  so  sloped  that  water  can  run  off,  and  the  ends 
should  be  stopped  with  paint.     It  is  well  to  paint  joints  before 
they  are  put  together. 

The  decay  of  timber  is  due  to  the  presence  and  action  of 
vegetable  growths  or  fungi,  the  spores  of  which  find  lodgment 
in  the  pores  of  the  wood,  but  require  air  and  moisture,  with  a 


MATERIALS.  21 

suitable  temperature,  for  their  germination  and  spread.  Hence 
if  timber  is  kept  perfectly  dry  it  will  last  indefinitely.  If  it  is 
entirely  immersed  in  water,  it  will  also  endure,  as  air  is  excluded. 
Moisture  may  be  excluded  from  an  exposed  surface  by  the  use 
of  paint.  Unseasoned  timber  painted,  or  placed  where  there  is 
no  circulation  of  air,  will  dry-rot  rapidly  in  the  interior  of  the 
stick;  but  the  exterior  shell  will  be  preserved,  since  it  dries  out 
or  seasons  to  a  little  depth  very  soon. 

The  worst  location  for  timber  is  at  or  near  the  ground  surface; 
it  is  then  continually  damp  and  rot  spreads  fast. 

31.  Preservation  of  Wood. — The  artificial  treatment  of  timber 
to  guard  against  decay  may  be  briefly  described  as  the  intro- 
duction into  the  pores  of  some  poison  or  antiseptic  to  prevent 
the  germination  of  the  spores;    such  treatment  is  efficacious  as 
long  as  the  substance  introduced  remains  in  the  wood.     Creosote 
is  the  best  of  preservatives  and  the  only  one  effective  against  sea- 
worms,  but  is  expensive. 

The  timber  is  placed  in  a  closed  tank  and  steam  is  admitted 
to  soften  the  cells.  After  some  time  the  steam  is  shut  off,  a 
partial  vacuum  is  formed,  and  the  preservative  fluid  is  run  in  and 
pressure  applied  to  force  the  liquid  into  the  pores  of  the  wood. 
As  steaming  injures  the  fibres,  treated  timber  is  weaker  than 
untreated. 

Burnettizing  is  the  name  given  to  treatment  with  zinc  chloride, 
a  comparatively  cheap  process,  applied  to  rail  way- ties  and  paving- 
blocks.  To  prevent  the  zinc  chloride  from  dissolving  out  in 
wet  situations,  tannin  has  been  added  after  the  zinc,  to  form 
with  the  vegetable  albumen  a  sort  of  artificial  leather,  plugging 
up  the  pores;  hence  the  name,  zinc-tannin  process.  For  bridge- 
timbers  burnettizing  makes  the  timber  unduly  brittle. 

As  the  outside  of  treated  timber  contains  most  of  the  preserv- 
ative, timber  should  be  framed  before  being  treated. 

32.  Strength   of   Timber. — The   properties   and   strength    of 
different  pieces  of  timber  of  the  same  species  are  very  variable. 
Seasoned  lumber  is  much  stronger  than  green,  and  of  two  pieces 
of  the  same  species  and  of  the  same  dryness,  the  heavier  is  the 
stronger,  while,  in  general,  heavy  woods  are  stronger  than  light. 


22 


STRUCTURAL  MECHANICS. 


Prudence  would  dictate  that  structures  should  be  designed  for 
the  strength  of  green  or  moderately  seasoned  timber  of  average 
quality.  As  the  common  woods  have  a  comparatively  low  re- 
sistance to  compression  across  the  grain,  particular  attention 
should  be  paid  to  providing  sufficient  bearing  area  where  a 
strut  or  post  abuts  on  the  side  of  another  timber.  An  indenta- 
tion of  the  wood  destroys  the  fibre  and  increases  the  liability  to 
decay,  if  the  timber  is  exposed  to  the  weather,  especially  under 
the  continued  working  produced  by  moving  loads. 

Average  breaking  stresses  of  some  American  timbers  as  found 
by  the  Forestry  Div.  of  the  U.  S.  Dept.  of  Agriculture  follow. 
The  results  are  for  well-seasoned  lumber,  12  per  cent,  moisture: 


Compression. 

Bending. 

Weight 

Shear 

per 

with 

Cu.  Ft. 

With 

Across 

Modulus 

Modulus 

Grain. 

Grain. 

Grain. 

Rupture. 

Elasticity. 

Long-leaf  pine  

38 

8,000 

1,260 

12,600 

2,070,000 

835 

Short  -leaf  pine  

32 

6,500 

1,050 

10,100 

1,680,000 

770 

White  pine  

24 

5>4oo 

700 

7,900 

1,390,000 

400 

Douglas  spruce  

32 

5>7°° 

800 

7,900 

1,680,000 

500 

White  oak  

5° 

8,500 

2,200 

13,100 

2,090,000 

1  ,000 

See  also  §  145. 

Timber  is  graded  or  classified  at  the  sawmills  according 
to  the  standard  rules  of  different  manufacturers'  associations, 
and  specifications  should  call  for  a  grade  of  lumber  agreeing 
with  the  classification  of  the  mills  of  the  region  where  the  lumber 
is  produced. 

33.  Iron  and  Carbon. — Cast  iron  and  steel  differ  from  each 
other  in  physical  qualities  on  account  of  the  different  percentages 
of  carbon  in  combination  with  the  iron.     Ordinary  cast  or  pig 
iron  contains   from  3^  to  4  per  cent,  of  carbon,  while  structural 
steel  contains  from  one  to  two  tenths  of  one  per  cent.    Wrought 
iron  and  steel  are  made  by  removing  the  metalloids  from  cast 
iron. 

34.  Cast  Iron. — Iron  ore,  which  is  an  oxide  of  iron,  is  put  in 
a  blast-furnace  together  with  limestone  and  coke.     Superheated 


MATERIALS.  23 

air  is  blown  in  at  the  bottom  of  the  furnace  and  the  burning  of 
the  coke  produces  a  high  temperature  and  removes  the  oxygen 
from  the  ore.  The  earthy  materials  in  the  ore  unite  with  the 
limestone  and  form  a  slag  which  floats  on  the  surface  of  the 
molten  metal  and  is  drawn  off  separately.  The  iron  is  run  off 
into  molds  and  forms  pig  iron. 

When  broken,  the  pig  is  seen  to  be  crystalline  and  its  color 
may  be  white  or  gray,  depending  upon  the  condition  of  the  carbon 
in  the  iron.  In  the  furnace  the  carbon  is  dissolved  in  the  bath, 
but  when  the  iron  solidifies  the  carbon  may  either  remain  in  solu- 
tion and  produce  white  iron,  or  part  of  the  carbon  may  precipitate 
in  the  form  of  scales  of  graphite  and  produce  gray  iron.  The 
condition  of  the  carbon  depends  partly  on  the  rate  of  cooling, 
but  more  on  the  other  elements  present.  White  iron  is  hard 
and  brittle;  gray  iron  is  tougher.  If  gray  iron  is  run  into  a  mold 
lined  with  iron,  it  is  chilled  from  the  surface  to  a  depth  of  one- 
half  to  three-fourths  of  an  inch;  that  is,  the  surface  is  turned 
to  white  iron  and  made  intensely  hard,  as  in  the  treads  of  car- 
wheels. 

Besides  the  carbon,  pig  iron  contains  more  or  less  silicon, 
usually,  from  one  to  two  per  cent.  It  tends  to  make  the  carbon 
take  the  graphitic  form.  Sulphur  makes  the  iron  hard  and 
brittle;  good  foundry  iron  should  not  contain  more  than  0.15 
of  one  per  cent.  Phosphorus  makes  molten  iron  fluid,  and 
irons  high  in  phosphorus  are  used  in  making  thin  and  intricate 
castings,  but  such  castings  are  very  brittle.  The  amount  of 
silicon  and  sulphur  in  pig  iron  can  be  controlled  by  the  furnace- 
man,  but  the  only  way  in  which  the  amount  of  phosphorus  can 
be  kept  down  is  by  using  pure  ores  and  fuel.  The  phosphorus 
in  pig  iron  to  be  used  for  making  steel  by  the  acid  Bessemer 
process  is  limited  to  one-tenth  of  one  per  cent.;  iron  fulfilling 
that  requirement  is  called  Bessemer  pig. 

The  tensile  strength  of  cast  iron  varies  from  15,000  to  35,000 
Ib.  and  the  compressive  strength  from  60,000  to  200,000  Ib.  per 
sq.  in.  The  modulus  of  elasticity  ranges  from  10  to  30  million 
pounds  per  square  inch.  For  ordinary  foundry  iron  the  tensile 
strength  is  usually  18,000  to  22,000  Ib.  and  the  modulus  of  elas- 


24  STRUCTURAL  MECHANICS. 

ticity  12  to  15  million.  As  cast  iron  is  brittle  and  likely  to  contain 
hidden  defects,  it  is  little  used  in  structural  work. 

35.  Wrought  Iron. — Wrought  iron  is  made  by  melting  pig 
iron,  and  cinder  which  contains  oxide  of  iron,  together  in  a  rever- 
beratory  furnace.  The  carbon  and  silicon  in  the  iron  unite  with 
the  oxygen  of  the  slag,  leaving  metallic  iron.  As  the  carbon  is 
removed  the  melting-point  rises,  and  since  the  temperature  of  the 
furnace  is  not  high  enough  to  keep  the  iron  fluid,  it  assumes  a  pasty 
condition.  The  semi-fluid  iron  is  collected  into  a  lump  by  the 
puddler  and  withdrawn  from  the  furnace.  It  is  then  much  like 
a  sponge;  the  particles  of  wrought  iron  have  adhered  to  one 
another,  but  each  particle  of  iron  is  more  or  less  coated  with  a 
thin  film  of  slag  and  oxide,  as  water  is  spread  through  the  pores 
of  a  partly  dry  sponge. 

The  lump  of  iron  is  put  into  a  squeezer,  and  the  fluid  slag 
and  oxide  drip  out  as  water  does  from  a  squeezed  sponge.  But, 
as  it  is  impracticable  to  squeeze  a  sponge  perfectly  dry,  so  it  is 
impracticable  to  squeeze  all  the  impurities  out  from  among  the 
particles  of  metallic  iron.  In  the  subsequent  processes  of  rolling 
and  re-rolling  each  globule  of  iron  is  elongated,  but  the  slag 
and  oxide  are  still  there;  so  that  the  rolled  bar  consists  of  a  col- 
lection of  threads  of  iron,  the  adhesion  of  which  to  each  other 
is  not  so  great  as  the  strength  of  the  threads. 

If  the  surface  of  an  iron  bar  is  planed  smooth  and  then  etched 
with  acid,  the  metal  is  dissolved  from  the  surface  and  the  black 
lines  of  impurities  are  left  distinctly  visible. 

That  wrought  iron  is  fibrous  is  then  an  accident  of  the  process 
of  manufacture,  and  does  not  add  to  its  strength.  If  these  im- 
purities had  not  been  in  the  iron  when  it  was  rolled  out,  it 
would  have  been  more  homogeneous  and  stronger.  The  fibrous 
fracture  of  a  bar  which  is  nicked  on  one  side  and  broken  by 
bending  is  not  especially  indicative  of  toughness;  for  soft  steel 
is  tough  and  ductile  without  being  fibrous. 

The  tensile  strength  of  wrought  iron  is  about  50,000  Ib.  per 
sq.  in.  and  its  modulus  of  elasticity  about  28,000,000.  Wrought 
iron  is  still  used  to  some  extent  when  it  is  necessary  to  weLl  the 
material,  but  soft  steel  has  largely  driven  it  from  the  market. 


MATERIALS.  25 

36.  Steel. — Structural  steel  is  made  from  pig  iron  by  burning 
out  the  metalloids. 

In  the  Bessemer  process  molten  pig  iron  is  run  into  a  converter 
and  cold  air  is  blown  through  the  metal  to  burn  out  the  carbon 
and  silicon.  The  combustion  generates  enough  heat  to  keep 
the  metal  fluid,  although  the  melting-point  rises.  When  the 
metal  is  free  from  carbon  and  silicon,  manganese  is  added  to 
remove  the  oxygen  dissolved  in  the  metal  and  to  make  the  steel 
tough  when  hot.  If  it  is  desired  to  add  carbon  to  make  a  stronger 
steel,  spiegel-iron  is  used,  which  is  a  pig  iron  containing  about 
12  per  cent,  manganese  and  4  or  5  per  cent,  carbon.  After  the 
manganese  is  added  the  metal  is  cast  into  an  ingot. 

In  America  the  acid  Bessemer  process  is  used  exclusively; 
that  is,  the  converters  are  lined  with  a  silicious  material  which 
necessitates  a  silicious  or  acid  slag,  and  consequently  there  is  no 
elimination  of  sulphur  and  phosphorus,  which  can  enter  a  basic 
slag  only.  This  process  is  the  cheapest  way  of  making  steel,  but 
the  product  is  not  so  uniform  as  that  of  the  open-hearth  furnace. 
Bessemer  steel  is  used  for  rails  and  for  buildings,  but  is  not  used 
for  first-class  structural  work. 

An  open-hearth  furnace  is  a  regenerative  furnace  having  a 
hearth  for  the  metal  exposed  to  the  flame.  Heat  is  generated 
by  burning  gas  which,  together  with  the  air-supply,  has  been 
heated  before  it  is  admitted  to  the  furnace.  An  intense  heat  is 
thus  produced  which  is  sufficient  to  keep  the  metal  fluid  after 
the  carbon  has  been  removed.  The  furnace  is  charged  with 
pig  iron  and  scrap  and,  after  the  charge  is  melted,  iron  ore  is 
added.  The  carbon  and  silicon  unite  with  the  oxygen  of  the  air 
and  of  the  ore.  When  the  required  composition  is  attained 
the  steel  is  drawn  off,  ferromanganese  is  thrown  into  the  ladle 
and  the  ingot  is  cast.  After  removal  from  the  molds  the  ingots 
are  heated  and  rolled  into  plates  or  shapes.  Steel  castings  are 
made  by  pouring  the  metal  from  the  furnace  or  the  converter 
into  molds. 

If  the  hearth  is  lined  with  sand,  the  slag  formed  during  the 
oxidation  is  silicious  or  acid  and  the  oxide  of  phosphorus,  which 
acts  as  an  acid,  reunites  with  the  iron.  If,  however,  the  hearth 


26 


STRUCTURAL  MECHANICS. 


is  lined  with  dolomite,  limestone  may  be  added  to  the  charge 
to  form  a  basic  slag  which  the  phosphorus  may  enter.  The 
sulphur  also  may  be  reduced  to  some  extent  by  the  basic  process, 
but  not  by  the  acid. 

The  tensile  strength  of  pure  iron  is  probably  about  40,000  Ib. 
per  sq.  in.,  but  the  presence  of  other  elements,  as  carbon  and 
phosphorus  in  small  quantities,  increases  its  strength  and  makes 
it  more  brittle.  The  element  which  increases  the  strength  most 
with  the  least  sacrifice  of  toughness  is  carbon,  and  it  is  the  element 
which  the  manufacturer  uses  to  give  strength.  In  structural 
steels  it  may  range  from  0.05  to  0.25  of  one  per  cent.  Phosphorus 
and  sulphur  are  kept  as  low  as  possible.  Phosphorus  makes 
the  steel  brittle  at  ordinary  temperatures,  while  sulphur  makes 
it  brittle  at  high  temperatures  and  likely  to  crack  when  rolled. 
Manganese  makes  the  steel  tough  while  hot.  It  ranges  from 
0.30  to  0.60  of  one  per  cent,  in  ordinary  structural  steels. 

The  strength  of  rolled  steel  depends  somewhat  upon  the 
thickness  of  the  material,  thin  plates  which  have  had  more  work 
done  upon  them  being  stronger.  The  softer  structural  steels  can 
be  welded  readily,  and  the  medium  with  care.  They  will  not 
temper.  The  modulus  of  elasticity  of  both  soft  and  medium 
steel  is  about  29,000,000  Ib.  per  sq.  in. 

37.  Classification  of  Steels. — The  following  classification  and 
requirements  are  taken  from  the  standard  specifications  of  the 
American  Society  for  Testing  Materials. 


Steel. 

Tensile 
Strength 
in 
1000  Lb. 

Elonga- 
tion in 
8  In. 
not  Less 
than 

P 

Acid 

P 
Basic 

S 

Mn 

not  more  than 

STRUCTURAL   STEEL  FOR   BRIDGES   AND   SHIPS. 


Rivet        

50  to  60 

26% 

o  08% 

O.O6% 

0.06% 

Soft    

52  to  62 

2S% 

o  08% 

0.06% 

0.06% 

Medium  

60  to  70 

22% 

0.08% 

0.06% 

0.06% 

STRUCTURAL   STEEL   FOR   BUILDINGS. 


Rivet  

50  to  60 

26% 

0.10% 

0.10% 

IVIcditm 

60  to  70 

22% 

0    10% 

O    10% 

MATERIALS. 

BOILER-PLATE   AND    RIVET-STEEL. 


Steel. 

Tensile 
Strength 
in 

Elonga- 
tion in 
8  In. 

P 

Acid 

P 

Basic 

S 

Mn 

1000  Lb. 

than 

nc 

t  more  ths 

n 

Extra-soft  
Fire-box 

45  to  55 
52  to  62 

28% 
26% 

0.04% 

O    O4.% 

0.04% 

O    OT.°7n 

o.a4% 

O    OA% 

o  .  30  to  o  .  50% 

Boiler 

c  c  to  6^ 

2^% 

o  06% 

O    Os% 

Steel  for  buildings  may  be  made  by  the  Bessemer  process;  the  other  two 
classes  must  be  made  by  the  open-hearth  process.  Test  specimens  of  rivet  and 
soft  steel  not  more  than  three-fourths  of  an  inch  thick  must  bend  cold  180°  flat 
without  fracture;  similar  specimens  of  medium  steel  must  bend  cold  180°  around 
a  diameter  equal  to  the  thickness  of  the  piece  without  fracture.  The  yield- 
point  must  not  be  less  than  one-half  the  ultimate  strength. 

This  classification  gives  a  good  idea  of  the  usual  require- 
ments for  steel,  although  some  engineers  prefer  a  grade  of 
structural  steel  midway  between  soft  and  medium,  that  is,  one 
having  a  strength  of  55  to  65  thousand  pounds  per  square  inch. 
Such  a  grade  is  recommended  by  the  American  Railway 
Engineering  and  Maintenance-of-Way  Association. 


38.  Work  of  Elongation. — It  is  seen  from  the  diagram  Fig.  i 
that  the  resistance  of  the  metal  per  square  inch  increases  as  the 
bar  draws  out,  and  diminishes  in  section  under  tension,  as  shown 
by  the  dotted  curve,  although  the  total  resistance  grows  less  near 
the  close  of  the  test,  as  shown  by  the  full  line.  As  a  small  increase 
in  the  amount  of  carbon  diminishes  the  elongation  and  reduction 
of  area,  it  is  possible  that  the  carbon  affects  the  apparent  ultimate 
strength  in  this  manner  (since  such  strength  is  computed  on  the 
square  inch  of  original  section),  and  not  by  actually  raising  the 
resisting  power  of  the  metal. 

Since  the  measure  of  the  work  done  in  stretching  a  bar  is  the 
product  of  one-half  the  force  by  the  stretch,  if  the  yield-point 
has  not  been  passed,  and,  for  values  beyond  that  point,  is  the 
area  below  the  curve  in  the  diagram,  limited  by  the  ordinate 
representing  the  maximum  force,  the  comparative  ability  of 
a  material  to  resist  live  load,  shock,  and  vibration  is  indicated 
by  this  area.  A  mild  steel  of  moderate  strength  may  thus  have 
greater  value  than  a  higher  carbon  steel  of  much  greater  tensile 
strength. 


28  STRUCTURAL  MECHANICS. 

39.  Tool-steel.  —  Tool-steel  as   well   as    spring-steel  of   good 
quality  is  made  by  melting  wrought  iron  or  steel  of  known  compo- 
sition in  a  crucible,  and  may  contain  from  one-half  to  one  per 
cent,  of  carbon.    When  heated  a  bright  red  and  quenched  in  water 
such  steel  becomes  very  hard  and  brittle  and  entirely  loses  the 
property  of  drawing  out;  but  if  it  is  subsequently  heated  to  a 
moderate    temperature    and    then    allowed    to    cool    slowly,    its 
strength  is  increased   and   its  brittleness   reduced,  while  it  still 
retains   more   or   less    of   its   hardness.     This   process   is   called 
tempering.     Springs  and  tools  are  tempered  before  being  used. 
Some  special  tool-steels  contain    tungsten  or  chromium,   which 
give  great  hardness  without  tempering. 

40.  Malleable    Iron:     Case-hardened    Iron. — There   are   two 
other  products  which  may  well  be  mentioned,  and  which  will  be 
seen  to  unite  or  fit  in  between  the  three  already  described.     The 
first  is  what  is  known  as  " malleable  cast  iron"  or  malleable  iron. 

Small  articles,  thin  and  of  irregular  shapes,  which  may  be 
more  readily  cast  than  forged  or  fashioned  by  a  machine,  and 
which  need  not  be  very  strong,  are  made  of  white  cast  iron,  and 
then  imbedded  in  a  substance  rich  in  oxygen,  as,  for  instance, 
powdered  red  hematite  iron  ore,  sealed  up  in  an  iron  box,  and 
heated  to  a  high  temperature  for  some  time.  The  oxygen  ab- 
stracts the  carbon  from  the  metal  to  a  slight  depth,  converting 
the  exterior  into  soft  iron,  while  the  carbon  in  the  interior  takes 
on  a  graphitic  form  with  an  increase  of  strength  and  diminution 
of  brittleness. 

The  second  product  is  case-hardened  iron.  An  article  fash- 
ioned of  wrought  iron  or  soft  steel  is  buried  in  powdered  charcoal 
and  heated.  The  exterior  absorbs  carbon  and  is  converted  into 
high  steel,  which  will  better  resist  wear  and  violence  than  will 
soft  iron.  The  Harvey  process  for  hardening  the  exterior  of 
steel  armor-plates  is  of  a  similar  nature. 

41.  Effect   of   Shearing  and   Punching. — As    was    shown    in 
§  15,  when  a  bar  of  steel  is  stressed  beyond  the  elastic  limit  and 
has  received  a  permanent  set,  a  higher  elastic  limit  is  established, 
but  the  percentage  of  elongation  is  much  reduced,  as  shown  by 
the  curve  E  F  N  of  Fig.  i.     The  steel  therefore  has  been  hardened 


MATERIALS.  29 

in  the  sense  that  its  ductility  has  been  lessened.  Examples  of 
this  hardening  are  seen  in  plates  which  are  rolled  cold  and  in 
drawn  wire.  Similarly  when  a  rivet -hole  is  punched  in  a  plate, 
the  metal  immediately  surrounding  the  hole  is  distorted  and 
hardened,  thus  reducing  the  ductility  of  the  plate  around  the 
hole.  If  the  strip  containing  a  punched  hole  is  loaded  beyond 
the  elastic  limit  of  the  plate,  the  metal  surrounding  the  hole, 
being  unable  to  stretch  as  much  as  the  rest,  is  unduly  stressed 
and  the  ultimate  strength  is  less  than  it  would  have  been  had 
the  stress  been  uniformly  distributed.  Experiments  show  that 
plates  with  punched  holes  are  weaker  than  those  with  drilled  holes. 

The  same  hardening  effect  is  produced  by  shearing,  or  cutting, 
a  plate.  When  a  bar  of  punched  or  sheared  steel  is  bent,  cracks 
form  at  the  hard  edges  and  spread  across  the  plate;  but  if  the 
holes  are  reamed  out  or  the  sheared  edges  are  planed  off  to  a 
small  depth,  the  hardened  metal  is  removed  and  the  bar  will 
bend  without  cracking.  Medium  steel,  especially  if  thick,  is 
injured  much  more  than  soft  steel  by  punching  and  shearing. 
Specifications  for  structural  work  frequently  require  rivet-holes  to 
be  reamed  to  a  diameter  three-sixteenths  of  an  inch  larger  than  the 
punch,  and  one-quarter  of  an  inch  to  be  planed  from  the  edges  of 
sheared  plates.  In  good  boiler- work  the  rivet-holes  are  drilled. 

The  ductility  of  steel  which  has  been  hardened  by  cold  work- 
ing can  be  restored  by  annealing,  that  is,  by  heating  to  a  red 
heat  and  then  cooling  slowly. 

42.  Building  Stone.  —  The  principal  building  stones  may 
be  grouped  as  granites,  limestones,  and  sandstones.  Granite 
consists  of  crystals  of  quartz,  felspar  and  mica  or  hornblende. 
It  is  very  strong  and  durable,  but  its  hardness  makes  it  difficult 
to  work.  Owing  to  its  composite  structure  it  does  not  resist 
fire  well.  Limestone  is  a  stratified  rock  of  which  carbonate  of 
lime  is  the  chief  ingredient.  When  limestone  consists  of  nearly 
pure  carbonate  and  is  of  good  color  and  texture,  it  is  called 
marble.  Sandstone  consists  of  grains  of  sand  cemented  together 
by  silica,  carbonate  of  lime,  iron  oxide,  or  clay.  If  the  cementing 
material  is  silica,  the  stone  is  very  hard  to  work.  Sandstone  is 
one  of  the  most  valuable  of  building  materials- 


30  STRUCTURAL  MECHANICS. 

Sound,  hard  stones  like  granite,  compact  limestone,  and  the 
better  grades  of  sandstone  are  sufficiently  strong  to  carry  any 
loads  brought  upon  them  in  ordinary  buildings ;  hence  the  question 
of  durability  rather  than  strength  is  the  governing  consideration 
in  selecting  a  good  building  stone.  The  only  sure  test  of  the 
ability  of  a  building  stone  to  resist  climatic  changes,  to  stand  the 
weather,  is  the  lapse  of  time.  Artificial  freezing  and  thawing 
of  a  small  specimen,  frequently  repeated,  will  give  indication  as 
to  durability. 

Stratified  stones  should  be  laid  on  their  natural  beds,  that  is,  so 
that  the  pressure  shall  come  practically  perpendicular  to  the  layers. 
They  are  much  stronger  in  such  a  position,  and  the  moisture 
which  porous  stones  absorb  from  the  rain  can  readily  dry  out. 
If  the  stones  are  set  on  edge,  the  moisture  is  retained  and,  in  the 
winter  season,  tends  to  dislodge  fragments  by  the  expansive  force 
exerted  when  it  freezes.  Some  sandstone  facings  rapidly  deterio- 
rate from  this  cause.  Crystals  of  iron  pyrites  occur  in  some  sand- 
stones and  unfit  them  for  use  in  the  face  of  walls.  The 
discoloration  resulting  from  their  oxidation,  and  the  local  break- 
ing of  the  stone  from  the  swelling  are  objectionable. 

The  modulus  of  elasticity  differs  greatly  for  different  stones. 
Limestones  and  granites  are  nearly  perfectly  elastic  for  all  work- 
ing loads,  but  sandstones  take  a  permanent  set  for  the  smallest 
loads.  Tests  of  American  building  stones  in  compression  made  at 
the  Watertown  Arsenal  give  values  of  5  to  10  million  for  granites 
and  marbles  and  i  to  3  million  for  sandstones.  The  weight  of 
granite  ranges  from  160  to  180,  of  limestone  and  marble  from 
150  to  170,  and  of  sandstone  from  130  to  150  pounds  per 
cubic  foot. 

43.  Masonry. — Most  masonry  consists  of  regularly  coursed 
stones  on  the  face,  with  a  backing  of  irregular-shaped  stones 
behind.  Stones  cut  to  regular  form  and  laid  in  courses  make 
ashlar  masonry,  if  the  stones  are  large  and  the  courses  continuous. 
When  the  stones  are  smaller,  and  the  courses  not  entirely  con- 
tinuous, or  sometimes  quite  irregular,  although  the  faces  are 
still  rectangular,  the  descriptive  name  is  somewhat  uncertain, 
as  block-in-course,  random  range,  etc.,  down  to  coursed  rubble, 


MATERIALS.  31 

where  the  end  joints  of  the  stones  are  not  perpendicular  to  the 
beds.  Rubble  masonry  denotes  that  class  where  the  stones  are 
of  irregular  shape,  and  fitted  together  without  cutting.  If  the 
face  of  the  stone  is  left  as  it  comes  from  the  quarry,  the  work  is 
called  quarry-faced  or  rock-faced.  The  kind  of  masonry  depends 
upon  the  beds  and  joints.  Walls  of  stone  buildings  have  only 
a  more  or  less  thin  facing  of  stone,  the  body  of  the  wall  being 
of  brick.  The  stone  facing  should  be  well  anchored  to  the  brick- 
work by  iron  straps. 

44.  Bricks. — Bricks  are  made  from  clay  which  may  be 
roughly  stated  to  be  silicate  of  alumina  (Al2O3,2SiO2,2H2O). 
The  clay  is  freed  from  pebbles,  mixed  with  water  in  a  pug-mill 
and  molded.  The  green  bricks  are  dried  in  the  air  and  then 
burned  in  kilns.  Pressed  bricks  are  pressed  after  drying.  They 
have  a  smooth  exterior,  are  denser  and  are  more  expensive  than 
common  bricks.  Paving-bricks  are  made  from  hard,  laminated, 
rock-like  clays  called  shales,  which  are  not  plastic  unless  pulver- 
ized and  mixed  with  water.  Paving- bricks  are  burned  to  incipient 
vitrification,  which  makes  them  extremely  hard.  Lime  and  iron 
in  clay  act  as  fluxes  and  make  the  clay  fusible;  fire-bricks  are 
therefore  made  from  clay  free  from  fluxes.  If  limestone  pebbles 
occur  in  a  brick-clay,  they  must  be  removed  or  they  will  form 
lumps  of  lime  after  burning,  and  when  wet  will  slake,  swell,  and 
break  the  bricks. 

The  red  color  of  common  bricks  is  an  accidental  character- 
istic, due  to  iron  in  the  clay.  Such  bricks  are  redder  the  harder 
they  are  burned,  finally,  in  some  cases,  turning  blue.  The  cream- 
colored  bricks  with  no  iron  may  be  just  as  strong  and  are  com- 
mon in  some  sections.  Soft,  underburned  bricks  are  very  porous, 
absorb  much  water,  and  cannot  be  used  on  the  outside  of  a 
wall,  especially  near  the  ground  line,  for  they  soon  disintegrate 
from  freezing.  Hard-burned  bricks  are  very  strong  and  satis- 
factory in  any  place;  they  can  safely  carry  six  or  eight  tons  to 
the  square  foot.  Bricks  differ  much  in  size  in  different  parts 
of  the  country.  A  good  brick  should  be  straight  and  sharp- 
edged,  reasonably  homogeneous  when  broken,  dense  and  heavy. 
Two  bricks  struck  together  should  give  a  ringing  sound. 


3  2  STRUCTURAL  MECHANICS. 

Sand  bricks  are  made  by  mixing  thoroughly  sand  with 
5  or  10  per  cent,  of  slaked  lime  and  sufficient  water  to  allow 
molding.  The  bricks  are  formed  under  very  great  pressure 
and  are  then  run  into  a  large  boiler  and  exposed  to  the  action  of 
steam  under  pressure  for  several  hours.  Some  chemical  reaction 
takes  place  between  the  silica  and  the  lime  under  the  conditions 
of  heat  and  moisture,  which  firmly  cements  the  particles  of  sand. 
Well-made  sand  bricks  have  a  crushing  strength  of  2,500  to  5,000 
Ib.  per  sq.  in.  They  are  denser  than  common  bricks  and  are 
very  regular  in  shape  and  size. 

45.  Lime. — Lime  for  use  in  ordinary  masonry  and  brick- 
work is  made  by  burning  limestone,  or  calcium  carbonate,  CaCOs, 
and  thus  driving  off  by  a  high  heat  the  carbon  dioxide  and  such 
water  as  the  stone  contains.  There  remains  the  quicklime  of  com- 
merce, CaO,  in  lumps  and  powder.  This  quicklime  has  a  great 
affinity  for  water  and  rapidly  takes  it  up  when  offered,  swelling 
greatly  and  falling  apart,  or  slaking,  into  a  fine,  dry,  white  powder, 
Ca(OH)2,  with  an  evolution  of  much  heat,  due  to  the  combina- 
tion of  the  lime  with  the  water.  The  use  of  more  water  produces 
a  paste,  and  the  addition  of  sand,  which  should  be  silicious, 
sharp  in  grain  and  clean,  makes  lime  mortar.  The  sand  is 
used  partly  for  economy,  partly  to  diminish  the  tendency  to  crack 
when  the  mortar  dries  and  hardens,  and  partly  to  increase  the 
crushing  strength.  The  proportion  is  usually  2  or  2j  parts  by 
measure  of  sand  to  one  of  slaked  lime  in  paste,  or  5  to  6  parts 
of  sand  to  one  of  unslaked  lime.  As  lime  tends  to  air-slake,  it 
should  be  used  when  recently  burned. 

Some  limes  slake  rapidly  and  completely;  other  limes  have 
lumps  which  slake  slowly  and  should  be  allowed  time  to  com- 
bine with  the  water.  It  is  generally  considered  that  lime  mortar 
improves  by  standing,  and  that  mortar  intended  for  plastering 
should  be  made  several  days  before  it  is  used.  Small  unslaked 
fragments  in  the  plaster  will  swell  later  and  crack  the  finished 
surface.  The  lime  paste  is  sometimes  strained  to  remove  such 
lumps. 

Lime  mortar  hardens  by  the  drying  out  of  part  of  the  water 
which  it  contains,  and  by  the  slow  absorption  of  carbon  dioxide 


MATERIALS.  33 

from  the  air.  It  thus  passes  back  by  degrees  to  a  crystallized  cal- 
cium carbonate  surrounding  the  particles  of  sand :  Ca(OH)2  +CO2 
=  CaCO3+H2O.  Dampness  of  the  mortar  is  favorable  to  the 
attainment  of  this  result,  and  the  mortar  in  a  brick  wall  which 
has  been  kept  damp  for  some  time  will  harden  better  than  where 
the  wall  is  dry.  Dry,  porous  bricks  absorb  rapidly,  and  almost 
completely,  the  water  from  the  mortar,  and  reduce  it  to  a  powder 
or  friable  mass  which  will  not  harden  satisfactorily.  Hence 
bricks  should  be  well  wetted  before  they  are  laid. 

Lime  mortar  in  the  interior  of  a  very  thick  wall  may  not 
harden  for  a  long  time,  if  at  all,  and  hence  should  not  be  used 
in  such  a  place.  Slaked  lime  placed  under  water  will  not  harden, 
as  may  be  proved  by  experiment.  In  both  cases  such  inaction 
is  due  to  the  exclusion  of  the  carbon  dioxide.  Lime  mortar 
should  never  be  used  in  wet  foundations. 

Plaster  for  interior  walls  is  lime  mortar.  Hair  is  added 
to  the  mortar  for  the  first  coat,  so  that  the  portion  which  is  forced 
through  the  spaces  between  the  laths  and  is  clinched  at  the  back 
may  have  sufficient  tenacity  to  hold  the  plaster  on  the  walls 
and  ceiling. 

46.  Natural  Cement. — Natural  cement  is  made  by  burning 
almost  to  vitrification  a  rock  which  contains  lime,  silica,  and 
alumina,  that  is,  one  which  may  be  considered  a  mixture  of  a 
limestone  and  a  clay  rock.  The  carbon  dioxide,  moisture,  and 
water  of  crystallization  are  expelled  by  burning.  The  hard 
fragments  must  then  be  ground  to  powder,  the  finer  the  better. 
If  the  rock  contains  the  several  ingredients  in  proper  proportions, 
the  addition  of  water  to  the  powder  makes  a  plastic  mass  which 
hardens  or  sets  by  crystallization.  This  setting  may  begin  in  a 
few  minutes  or  half  an  hour.  The  hardening,  the  tensile  and 
compressive  strengths  increase  rapidly  at  first,  and  at  a  decreasing 
rate  for  months. 

As  access  of  air  is  not  required  for  the  setting  of  cement, 
the  reaction  taking  place  when  water  is  added  to  the  dry  powder, 
cement  mortar  is  used  invariably  under  water  and  in  wet  place:-. 
It  makes  stronger  work  than  lime  mortar,  and  is  generally  ULC  1 
by  engineers  for  stone  masonry.  Its  greater  cost  than  that  of 


34  STRUCTURAL  MECHANICS. 

lime  is  due  to  the  necessity  of  grinding  the  hard  clinker;  while 
lime  falls  to  powder  when  wet.  The  proportion  of  sand  is  i,  2, 
or  3  to  one  of  cement,  according  to  the  strength  desired,  2  to  i 
being  a  common  ratio  for  good  work.  The  sand  and  cement  are 
mixed  dry  and  then  wetted,  in  small  quantities,  to  be  used  at  once. 
The  addition  of  brick-dust  from  well-burned  bricks  to  lime 
mortar  will  make  the  latter  act  somewhat  like  cement,  or  become 
hydraulic,  as  it  is  called.  Volcanic  earth  has  been  used  in  the 
same  way. 

47.  Portland  Cement. — If  the  statement  made  as  to  the  com- 
position of  cement  is  correct,  it  should  be  possible  to  make  a 
mixture  of  chalk,  lime  or  marl,   and  clay  in  proper  proportions 
for  cement,  and  the  product  ought  to  be  more  uniform  in  com- 
position and  characteristics   than  that  from  the  natural  rock. 
Such  is  the  case,  and  in  practice  about  three  parts  of  carbonate 
of   lime  are  intimately  mixed  with  one  part  of  clay  and  burned 
in  kilns.     During  the  burning  the  combined  water  and  carbon 
dioxide  are  driven  off  and  various  compounds  are  formed  of  which 
tricalcium   silicate  (3CaO,SiO2)  is  the  most    important,  as  it  is 
the  principal  active  element  and  constitutes  the  greater  part  of 
hydraulic  cements.     The  resulting  clinker  is  ground  to  an  impal- 
pable powder  which  forms  the*  Portland  cement  of  commerce. 
Fine  grinding  is  essential,  as  it  has  been  shown  that  the  coarser 
particles  of  the  cement  are  nearly  inert. 

Upon  mixing  the  cement  with  water,  the  soluble  salts  dis- 
solve and  crystallize,  that  is,  the  cement  sets.  The  water  does 
not  dry  out  during  hardening  as  in  lime  mortar,  but  combines 
with  some  of  the  salts  as  water  of  crystallization.  This  crystal- 
lization takes  place  more  slowly  in  the  Portland  than  in  the 
natural  cements,  but  after  the  Portland  cement  has  set  it  is  much 
harder  and  stronger  than  natural  cement.  The  slower-setting 
cement  mortars  are  likely  to  show  a  greater  strength  some  months 
or  years  after  use  than  do  the  quick-setting  ones,  which  attain 
considerable  strength  very  soon,  but  afterwards  gain  but  little. 

48.  Cement   Specifications. — The  following  specifications   of 
cement  are  reasonable: 


MATERIALS.  35 


NATURAL   CEMENT. 

Specific  gravity:  not  less  than  2.8. 

Fineness:  90  per  cent,  to  pass  a  sieve  of  10,000  meshes  per 
square  inch. 

Setting:  initial  set  in  not  less  than  fifteen  minutes;  final  set 
in  not  more  than  four  hours. 

Soundness:  thin  pats  of  neat  cement  kept  in  air  or  in  water 
shall  remain  sound  and  show  no  cracks. 

Tensile  strength:  briquettes  one  inch  square  in  cross-section 
shall  develop  after  setting  one  day  in  air  and  the  remaining  time 
in  water : 

Neat 7  days 100  Ib. 

"    28      "    200  Ib. 

i  cement,  i  sand . .   7      "    60  Ib. 

"      ..28     " .i5olb. 


PORTLAND   CEMENT. 

Specific  gravity:  not  less  than  3.10. 

Fineness:  92  per  cent,  to  pass  a  sieve  of  10,000  meshes  per 
square  inch. 

Setting:  initial  set  in  not  less  than  thirty  minutes ;  final  set  in 
not  more  than  ten  hours.  (If  a  quick-setting  cement  is  desired 
for  special  work,  the  time  of  setting  may  be  shortened  and  the 
requirements  for  tensile  strength  reduced.) 

Soundness:  a  thin  pat  of  neat  cement  kept  in  air  28  days 
shall  not  crack;  another  pat  allowed  to  set  and  then  boiled  for 
five  hours  shall  remain  sound. 

Tensile  strength:  briquettes,  as  for  natural  cement: 


Neat 

7  davs 

450  Ib. 

28     " 

550  Ib. 

i  cement 

3  sand       7     '  ' 

.     150  Ib. 

11      ..28     " 

.  .  200  Ib. 

49.  Concrete. — Concrete  is  a  mixture  of  cement  mortar  (cement 
and  sand)  with  gravel  and  broken  stone,  the  materials  being  so 
proportioned  and  thoroughly  mixed  that  the  gravel  fills  the 
spaces  among  the  broken  stone;  the  sand  fills  the  spaces  in  the 
gravel;  and  the  cement  is  rather  more  than  sufficient  to  fill  the 


36  STRUCTURAL  MECHANICS. 

interstices  of  the  sand,  coating  all,  and  cementing  the  mass  into 
a  solid  which  possesses  in  time  as  much  strength  as  many  rocks. 
It  is  used  in  foundations,  floors,  walls,  and  for  complete  structures. 
The  broken  stone  is  usually  required  to  be  small  enough  to  pass 
through  a  2 -in.  or  2|-in.  ring.  The  stone  is  sometimes  omitted. 

To  ascertain  the  proportions  for  mixing,  fill'a  box  or  barrel 
with  broken  stone  shaken  down,  and  count  the  buckets  of  water 
required  to  fill  the  spaces;  then  empty  the  barrel,  put  in  the 
above  number  of  buckets  of  gravel,  and  count  the  buckets  of 
water  needed  to  fill  the  interstices  of  the  gravel;  repeat  the 
operation  with  that  number  of  buckets  of  sand,  and  use  an  amount 
of  cement  a  little  more  than  sufficient  to  fill  the  spaces  in  the  sand. 
If  the  gravel  is  sandy,  screen  it  before  using,  in  order  to  keep 
the  proportions  true.  A  very  common  rule  for  mixing  is  one 
part  cement,  three  parts  sand,  and  five  parts  broken  stone  or 
pebbles,  all  by  measure. 

The  ingredients  are  mixed  dry,  then  water  is  added  and  the 
mass  is  mixed  again,  after  which  it  is  deposited  in  forms  in  layers 
6  or  8  inches  thick.  Experience  has  shown  that  a  mixture  wet 
enough  to  flow  makes  a  denser  concrete  than  a  dry  mixture, 
especially  if  the  mass  cannot  be  thoroughly  tamped. 

50.  Paint. — When  a  film  of  linseed- oil,  which  is  pressed  from 
flaxseed,  is  spread  on  a  surface  it  slowly  becomes  solid,  tough, 
and  leathery  by  the  absorption  of  oxygen  from  the  air.  In  order 
that  the  film  may  solidify  more  rapidly  the  raw  oil  may  be  pre- 
pared by  heating  and  adding  driers,  oxides  of  lead  and  man- 
ganese, which  aid  the  oxidation;  oil  treated  in  this  way  is  called 
boiled  oil.  Driers  should  be  used  sparingly,  as  they  lessen  the 
durability  of  paint.  An  oil-film  is  somewhat  porous  and  rather 
soft,  hence  its  protective  and  wearing  qualities  can  be  improved 
by  the  addition  of  some  finely  ground  pigment  to  fill  the  pores 
and  make  the  film  harder  and  thicker.  Most  pigments  are  inert. 
Paint,  then,  consists  of  linseed-oil,  a  pigment,  and  a  drier.  Varnish 
is  sometimes  added  to  make  the  paint  glossy  and  harder,  or  tur- 
pentine may  be  used  to  thin  it. 

Varnishes  are  made  by  melting  resin  (resins  are  vegetable 
gums,  either  fossil  or  recent),  combining  it  with  linseed-oil,  and 


MATERIALS.  37 

thinning  with  turpentine.  They  harden  by  the  evaporation  of 
the  turpentine  and  the  oxidation  of  the  oil  and  resin.  The 
addition  of  a  pigment  to  varnish  makes  enamel-  or  varnish - 
paint. 

For  painting  on  wood  white  lead,  the  carbonate,  and  white 
zinc,  an  oxide,  are  pigments  extensively  used.  Iron  oxide  is 
largely  used  on  both  wood  and  steel.  Red  lead,  an  oxide,  and 
graphite  are  pigments  used  on  steel.  Red  lead  acts  differently 
from  other  pigments  in  that  it  unites  with  the  oil,  and  the  mixture 
hardens  even  if  the  air  is  excluded,  so  thai  red-lead  paint  must 
be  mixed  as  used.  Lampblack  is  often  mixed  with  other  pig- 
ments to  advantage,  or  it  is  sometimes  used  alone. 

As  paint  is  used  to  form  a  protective  coating,  it  should  not 
be  brushed  out  too  thin,  but  as  heavy  a  coat  as  will  dry  uniformly 
should  be  applied.  Wood  should  be  given  a  priming  coat  of 
raw  linseed-oil,  so  that  the  wood  shall  not  absorb  the  oil  from  the 
first  coat  of  paint  and  leave  the  pigment  without  binder.  In 
applying  paint  to  steel-work  it  is  essential  for  good  work  that 
the  paint  be  spread  on  the  clean,  bright  metal.  Rust  and  mill- 
scale  must  be  removed  before  painting  if  the  coat  is  expected 
to  last.  As  mill-scale  can  be  removed  only  by  the  sand-blast 
or  by  pickling  in  acid,  steel  is  seldom  thoroughly  cleaned  in 
practice.  If  paint  is  applied  to  rusty  iron,  the  rusting  will  go  on 
progressively  under  the  paint.  Painting  should  never  be  done  in 
wet  or  frosty  weather. 


CHAPTER  III. 

BEAMS. 

51.  Beams:    Reactions. — A  beam  may  be  defined  to  be  a 
piece  of  a  structure,  or  the  structure  itself  as  a  whole,  subjected 
to  transverse  forces  and  bent  by  them.     If  the  given  forces  do 
not  act  at  right  angles  to  the  axis  or  centre  line  of  the  piece,  their 
components  in  the  direction  of  the  axis  cause  tension  or  com- 
pression, to  be  found  separately  and  provided  for;   the  normal  or 
transverse  components  alone  produce  the  beam  action  or  bending. 

As  all  trusses  are  skeleton  beams,  the  same  general  principles 
apply  to  their  analysis,  and  a  careful  study  of  beams  will  throw 
much  light  on  truss  action. 

Certain  forces  are  usually  given  in  amount  and  location  on 
a  beam  or  assumed.  Such  are  the  loads  concentrated  at  points 
or  distributed  over  given  distances,  and  due  to  the  action  of 
gravity;  the  pressure  arising  from  wind,  water,  or  earth;  or  the 
action  of  other  abutting  pieces. 

It  is  necessary,  in  the  first  place,  to  satisfy  the  requirements 
of  equilibrium,  that  the  sum  of  the  transverse  forces  shall  equal 
zero  and  that  the  sum  of  their  moments  about  any  point  shall 
also  equal  zero.  This  result  is  accomplished  by  finding  the 
magnitudes  and  direction  of  the  forces  required  at  certain  given 
points,  called  the  points  of  support,  to  produce  equilibrium. 
The  supporting  forces  or  reactions,  exerted  by  the  points  of  support 
against  the  beam,  are  two  or  more,  except  in  the  rare  case  where 
the  beam  is  exactly  balanced  on  one  point  of  support.  For  cases 
where  the  reactions  number  more  than  two,  see  §  109. 

52.  Beam    Supported    at   Two     Points.      Reactions.  —  The 
simplest  and  most  generally  applicable  method  for  finding  one 

38 


BEAMS.  39 

of  the  two  unknown  reactions  is  to  find  the  sum  of  the  moments 
of  the  given  forces  about  one  of  the  points  of  support,  and  to 
equate  this  sum  with  the  moment  of  the  other  reaction  about  the 
same  point  of  support.  Hence,  divide  the  sum  of  the  moments 
of  the  given  external  forces  about  one  of  the  points  of  support 
by  the  distance  between  the  two  points  of  support,  usually  called 
the  span,  to  find  the  reaction  at  the  other  point  of  support.  The 
direction  of  this  reaction  is  determined  by  the  sign  of  its  moment, 
as  required  for  equilibrium.  The  amount  of  the  other  reaction 
is  usually  obtained  by  subtracting  the  one  first  found  from  the 
total  given  load. 

W*~{QQ  ,Tr=7dO  TF=150  -600=P2, 

J<"Q-»g  20-       -~^          I£~~10~~^.D  A      B! 


Fig.  3  'XFig.4  Fig.  5      r' 

Thus,  in  the  three  cases  sketched,  pi  =  W~T^\  P2=W-Pi. 

Examples.— Fig.  3.     If  W=  500  lb.,  A  B  =  3o  ft.,  and  B  C  =  18  ft.; 
PI  =  —      -=300  lb.,  P2=  500— 300=200  lb. 

Fig.  4.     If  TF=75o  lb.,  AB  =  2o  ft.,  and  AC  =  5  ft.,  P1  =  75°'2^  = 

20 
937J  lb.,  and  ^3=750-937^-187^  lb. 

Fig.  5.     If  PF=i5olb.,  AC=2oft,  and  AB  =  5  ft,  P1  =  I^^1= 

750  lb.,  and  P2=  150-  750=  -600  lb.  Note  the  magnitude  of  PI  and 
P2  as  compared  with  W  when  the  distance  between  PI  and  P2  is  small. 
Such  is  often  the  case  when  the  beam  is  built  into  a  wall. 

Where  the  load  is  distributed  at  a  known  rate  over  a  certain 
length  of  the  beam,  the  resultant  load  and  the  distance  from  its 
point  of  application  to  the  point  of  support  may  be  conveniently 
u^ed. 


Fig.  6. 

Example.— Fig.  6.     If  AB  =  4o  ft,  AD  =  8  ft.,  DE=i6  ft.,  and 
the  load  on  DE  is  200  lb.   per  ft.,  TF=3,2oo  lb.,  and   CB  =  24  ft. 


40  STRUCTURAL  MECHANICS. 

Therefore  PI  =  ^—     —=1,920  lb.,  and  P2= 3, 200— 1,920=1,280  lb. 
40 

If  several  weights  are  given  in  position  and  magnitude,  the 
same  process  for  finding  the  reactions,  or  forces  exerted  by  the 
points  of  support  against  the  beam,  is  applicable. 

Examples. — In  Fig.  7,  PI  =  (100-18  +  200-  i6+i50'i3  +  3oo-n  + 
50-8+80.0)4-16  =  6651  lb.  P2= 880- 665!=  214!  lb.  The  work  can 


li'200 


be  checked  by  taking  moments  about  A  to  find  P2,  the  moment  100-2 
then  being  negative. 

If  the  depth  of  water  against  a  bulkhead,  Fig.  8,  is  9  ft.,  and  the 
distance  between  A  and  B,  the  points  of  support,  is  6  ft.,  A  being  at 
the  bottom,  the  unit  water  pressure  at  A  will  be  9X62.5  =  562.5  lb. 
which  may  be  represented  by  A  D,  and  at  other  points  will  vary  with 
the  depth  below  the  surface,  or  as  the  ordinates  from  E  A  to  the  inclined 
line  E  D.  Hence  the  total  pressure  on  E  A,  for  a  strip  i  ft.  in  hori- 
zontal width,  will  be  562. 5X9^-2  =  2, 531^  lb.,  and  the  resultant  pres- 
sure will  act  at  C,  distant  J  A  E,  or  3  ft.  from  A.  P2=  2, 53 1^X3-^6= 
1,265.6  lb.,  and  PI  =  2, 531. 2— 1,265.6=1,265.6  lb.,  a  result  that  might 
have  been  anticipated,  from  the  fact  that  the  resultant  pressure  here 
passes  midway  between  A  and  B. 

Let  1,000  lb.  be  the  weight  of  pulley  and  shaft  attached  by  a  hanger 
to  the  points  D  and  E,  Fig.  9.  Let  the  beam  A  B=  10  ft.,  A  D  =  4  ft., 
D  E  =  4  ft.,  E  B=2  ft.;  and  let  C  be  2  ft.  away  from  the  beam.  As 
the  beam  is  horizontal,  PI  =  1,000X4^-10= 400  lb. ;  P2=  1,000—400= 
600  lb.,  and  both  act  upwards.  The  1,000  lb.  at  C  causes  two  vertical 
downward  forces  on  the  beam,  each  500  lb.,  at  D  and  E.  There  is  also 
compression  of  500  lb.  in  D  E. 

When  the  beam  is  vertical,  Fig.  10,  by  moments,  as  before,  about 
B,  PI  =1,000 -2 -MO  =200  lb.  at  A  acting  to  the  left,  being  tension  or 
a  negative  reaction.  By  moments  about  A,  P2=i,ooo-2-Mo=2oo 
at  B,  acting  to  the  right.  Or  Pi  +  P2=o;  .  •  .  PI=  -P2.  By  similar 
moments,  the  1,000  lb.  at  C  causes  two  equal  and  opposite  horizontal 
forces  on  the  beam  at  D  and  E,  of  500  lb.  each,  that  at  D  being  ten- 
sion on  "the  connection,  or  acting  towards  the  right,  and  that  at  E , 
acting  in  the  opposite  direction.  These  two  forces  make  a  couple 


BEAMS.  41 

balanced  by  the  couple  P\P<i.  The  weight  1,000  Ib.  multiplied  by 
its  arm  2  ft.  is  balanced  by  the  opposing  horizontal  forces  at  D  and  E, 
4  ft.  apart.  There  remains  a  vertical  force  of  1,000  Ib.  in  A  B,  which 
may  all  be  resisted  by  the  point  B,  when  the  compression  in  D  E  = 
500  Ib.  and  in  E  B=  1,000  Ib.;  or  all  by  the  point  A,  when  the  tension 
in  D  £=500  Ib.  and  in  D  A=  1,000  Ib.;  or  part  may  be  resisted  at  A, 
and  the  rest  at  B,  the  distribution  being  uncertain.  This  longitudinal 
force  may  be  disregarded  in  discussing  the  beam,  as  may  the  tension 
or  compression  in  the  hanger  arms  themselves. 

53.  Bending  Moments. — If  an  imaginary  plane  of  section 
is  passed  through  any  point  in  a  beam,  the  sum  of  the  moments 
of  all  the  external  forces  on  one  side  of  that  section,  taken  about 
a  point  in  the  section,  must  be  exactly  equal  and  opposite  to 
the  sum  of  the  moments  of  all  the  external  forces  on  the  other 
side  of  that  section,  taken  about  the  same  point.  If  not,  the  beam 
would  revolve  in  the  plane  of  the  forces.  The  moment  on  the 
left  side  of  the  section  tends  to  make  that  portion  of  the  beam 
rotate  in  one  direction  about  the  point  of  section,  and  the  equal 
moment  on  the  right  side  of  the  section  tends  to  make  the  right 
segment  rotate  in  the  opposite  direction.  These  two  moments 
cause  resistances  in  the  interior  of  the  beam  at  the  section  (which 
stresses  will  be  discussed  under  resisting  moment],  with  the  result 
that  the  beam  is  bent  to  a  slight  degree.  Either  resultant  moment 
on  one  side  of  a  plane  of  section,  about  the  section,  is  called  the 
bending  moment  at  that  point,  usually  denoted  by  M ,  and  is  con- 
sidered positive  when  it  makes  the  beam  concave  on  the  upper 
side.  Ordinary  beams,  supported  at  the  ends  and  carrying 
loads,  have  positive  bending  moments. 

If  upward  reactions  are  positive,  weights  must  be  taken  as 
negative  and  their  sign  regarded  in  writing  moments. 

Examples.— Section  at  D,  Fig.  3,  10  ft.  from  B.  On  the  left  of  D, 
and  about  D,  PI  (=300) -20— 500-8=2,000  ft.-lb.,  positive  bending 
moment  at  D.  Or,  about  D,  on  the  right  side  of  the  section; 
P2(=  200) -10=  2,000  ft.-lb.,  positive  bending  moment  at  D.  Usually 
compute  the  simpler  one. 

Section  at  A,  Fig.  4,  W-C  A=  —  750-5=  -3, 75°  ft.-lb.  negative 
bending  moment  at  A,  tending  to  make  the  beam  convex  on  the  upper 
side.  At  D,  10  ft.  from  B,  M=—P2-io=- 187^-10=  — 1,875  ft.-lb., 
negative  because  PI  is  negative.  At  A,  taking  moments  on  the  right 


42  STRUCTURAL  MECHANICS. 

of  and  about  A,  M=  — 1874-20=—  3,750  ft.-lb.,  as  first  obtained. 
This  beam  has  negative  bending  moments  at  all  points. 

In  Fig.  5,  M  at  D  is— 150- 10=  —  1,500  ft.-lb.  It  is  evident  that 
the  bending  moments  at  all  points  between  C  and  A  can  be  found 
found  without  knowing  the  reactions.  If  this  beam  is  built  into  a 
wall,  the  points  of  application  of  PI  and  P2  are  uncertain,  as  the  pres- 
Slires  at  A  and  B  are  distributed  over  more  or  less  of  the  distance  that 
the  beam  is  embedded.  The  maximum  M  is  at  A,  and  is  —150-20= 

—  3,000  ft.-lb.     It  is  evident  that  the  longer  A  B  is,  the  smaller  the  reac- 
tions are,  and  hence  the  greater  the  security. 

In  Fig.  6,  the  bending  moment  at  C  will.be  PI -AC  — weight  on 
D  C-^D  0=1,920- 16-200-8-4=24,320  ft.-lb.  At  E,  M=i, 280-16= 
20,480  ft.-lb. 

In  Fig.  7,  the  bending  moments  at  the  several  points  of  application 
of  the  weights,  taking  moments  of  all  the  external  forces  on  the  left  of 
each  section  about  the  section,  will  be — 

At  C,  M=  — 100-0=0. 

At  A,  M=  —  loo- 2=  —  200  ft.-lb. 

AtD,  Jf=- loo- 5 +  (665!- 200) -3  =  896!  ft.-lb. 

At  E,  M=  -  loo-  7  +  465!-  5- 150- 2  =  ^3284-  ft.-lb. 

At  F,  M=-ioo-io+465f-8-i5o-5-3oo-3  =  i,o75  ft.-lb. 

And,  at  B,  M  will  be  zero.     M  max.  occurs  at  E. 

Do  not  assume  that  the  maximum  bending  moment  will 
be  found  at  the  point  of  application  of  the  resultant  of  the  load. 
The  method  for  finding  the  point  or  points  of  maximum  bending 
moment  will  be  shown  later. 

The  moments  on  the  right  portion  of  the  beam  may  be  more 
easily  found  by  taking  moments  on  the  right  side  of  any  section. 
Thus  at  F,  M  =  (P2-8o)-8  =  (2i4f-8o)-8  =  i,o75  ft.-lb.  Find 
the  bending  moment  at  the  middle  of  E  F.  1,201  A  ft.-lb. 

In  Fig.  8,  the  bending  moment  at  section  C  of  the  piece  A  E  may 
be  found  by  considering  the  portion  above  C.  As  the  unit  pressure 
at  C  is  6X62^  lb.  =  375  lb.  per  sq.  ft.,  M  at  O  P2(=i,265.6)-3 

—  (375X6-5-2) -6-*-3  =  1,546.8     ft.-lb.       At      the     section     B,     M= 
- (3X62^X3 -^-2)Xi  =  - 281 J  ft.-lb. 

In  Fig.  9,  as  PI  =  400  lb.,  P2=6oo  lb.,  vertical  forces  at  D  and  E 
are  each  500  lb.;  M  at  D=i,6oo  ft.-lb.;  M  at  E=  1,200  ft.-lb. 

In  Fig.  10,  as  P\=  —  200  lb.=  —  P2,  and  the  horizontal  forces  at  D 
and  E  are  ±  500  lb.;  M  at  D=-8oo  ft.-lb.;  M  at  £=+400  ft.-lb. 
The  beam  will  be  concave  on  the  left  side  at  D  and  convex  at  E.,  The 
curvature  must  change  between  D  and  E,  where  M=o.  Let  this 
point  be  distant  x  from  B.  Then  '200 •#— 500(^—2)  =  o;  .'.  x= 
3*  ^ 


BEAMS  43 

The  curved  piece  A  B,  Fig.  u,  with  equal  and  opposite  forces 
applied  in  the  line  connecting  its  ends,  will  experience  a  bending 
moment  at  any  point  D,  equal  to  P-CD,  this  ordinate  being 
perpendicular  to  the  chord. 

54.  Shearing  Forces. — In  Fig.  3,  of  the  500  Ib.  at  C,  300  Ib. 
goes  to  A  and  200  Ib.  to  B.     Any  vertical  section  between  A  and  C 
must  therefore  have  300  Ib.  acting  vertically  in  it.     On  the  left 
of  such  a  section  there  will  be  300  Ib.  from  PI  acting  upwards, 
and  on  the  right  of  the  same  section  there  will  be  300  Ib.,  coming 
from  W,  acting  downwards.     These  two  forces,  acting  in  opposite 
directions  on  the  two  sides  of  the  imaginary  section,  tend  to  cut 
the  beam  off,  as  would  a  pair  of  shears,  and  either  of  these  two 
opposite  forces  is  called  the  shearing  force  at  the  section,  or  simply 
the  shear.     When  acting  upwards  on  the  left  side  of  the  section 
(and  downwards  on  the  right  side),  it  is  called  positive  shear. 
When  the  reverse  is  the  case  the  shear  will  be  negative. 

Examples. — In  Fig.  7,  where  a  number  of  forces  are  applied  to  a 
beam,  there  must  be  found  at  any  section  between  C  and  A  a  shear 
of  — 100  Ib.;  between  A  and  D  the  shear  will  be  — 100  +  665!— 200= 
+  365!  Ib.;  between  D  and  E  the  shear  will  decrease  to  3651—150= 
215!  Ib.;  on  passing  E  the  shear  will  change  sign,  being  2151—300= 
—  84!  Ib.;  between  F  and  B  it  will  be  — 84!— 50=  - 134!  Ib.;  and 
on  passing  B,  it  becomes  zero,  a  check  on  the  accuracy  of  the  several 
calculations. 

In  Fig.  8,  the  shear  just  above  the  support  B  =  3X62|X3^2  = 
281^  Ib.;  just  below  the  point  B  the  shear  is  281^—1,265.6= —984.4  Ib.; 
and  just  above  A  it  is  1,265.6  Ib.  The  signs  used  imply  that  the  left 
side  of  A  E  corresponds  to  the  upper  side  of  an  ordinary  beam.  As 
the  shear  is  positive  above  A  and  negative  below  B,  it  changes  sign 
at  some  intermediate  point.  Find  that  point. 

In  Fig.  9,  the  shear  anywhere  between  A  and'  D  is  +400  Ib.;  at 
all  points  between  D  and  E  it  is  400— 500=  — 100  Ib.;  and  between 
E  and  B  is  —600  Ib.  The  shear  changes  sign  at  D. 

In  Fig.  10,  the  shear  on  any  horizontal  plane  of  section  between 
B  and  E  is  —200  Ib.;  betwen  E  and  D  is  —200+500= +300  Ib.; 
and  between  D  and  A  is  +300— 500=  — 200  Ib.  The  shear  changes 
sign  at  both  E  and  D. 

55.  Summary. — To    repeat: — The    shearing    jorce    at    any 
normal  section  of  a  beam  may  therefore  be  denned  to  be  the 
algebraic  sum  of  all  the  transverse  forces  on  one  side  of  the  section. 


44  STRUCTURAL   MECHANICS. 

When  this  sum  or  resulting  force  acts  upward  on  the  left  of  the 
section,  call  it  positive;   when  downward,  negative. 

The  bending  moment  at  any  right  or  normal  section  of  a  beam 
may  be  stated  to  be  the  algebraic  sum  oj  the  moments  oj  all  the 
transverse  forces  on  one  side  oj  the  section,  taken  about  the  centre 
of  gravity  oj  the  section  as  axis.  When  this  sum  or  resulting 
moment  is  right-handed  or  clockwise  on  the  left  of  and  about 
the  section,  call  it  positive.  A  positive  moment  tends  to  make 
the  beam  concave  on  what  is  usually  the  upper  side. 

By  a  proposition  in  mechanics,  any  force  which  acts  at  a 
given  distance  from  a  given  point  is  equivalent  to  the  same  force 
at  the  point  and  a  moment  made  up  of  the  force  and  the  perpen- 
dicular from  the  point  to  the  line  of  action  of  the  force.  Then 
in  Fig.  7,  if  a  section  plane  is  passed  anywhere,  as  between  D 
and  E,  the  resultant  force  on  the  left,  which  is  the  algebraic  sum 
of  the  given  forces  on  the  left  of  the  section,  is  the  shear  at  the 
section;  and  this  resultant,  multiplied  by  its  arm  or  distance 
from  the  point  in  D  E,  giving  a  moment  which  is  the  algebraic 
sum  of  the  moments  of  the  several  forces  on  the  left  of  and  about 
the  point,  is  the  bending  moment  at  the  section. 

It  is  also  evident  that  the  resulting  action  at  any  section  is 
the  sum  of  the  several  component  actions;  and  hence  that  dif- 
ferent loads  may  be  discussed  separately  and  their  effects  at 
any  point  added  algebraically,  if  they  can  occur  simultaneously. 
Thus  the  shears  and  bending  moments  arising  from  the  weight 
of  a  beam  itself  may  be  determined,  and  to  them  may  be  added 
the  shears  and  bending  moments  at  the  same  points  from  other 
weights  imposed  on  the  beam. 

The  numerous  examples  already  given  show  that  formulas 
are  not  needed  for  solving  problems  in  beams,  and  the  student 
will  do  well  to  accustom  himself  to  using  the  data  directly. 
Formulas,  however,  will  now  be  derived,  which  will  sometimes 
be  convenient  for  use,  and  from  which  may  be  deduced  certain 
serviceable  relationships. 

56.  Bending  Moment  a  Maximum  where  the  Shear  Changes 
Sign. — If  a  beam  weighing  w  per  unit  of  length  is  supported 
at  each  end  and  carries  a  system  of  loads  any  one  of  which  is 


BEAMS.  45 

distant  a  from  the  left  support,  the  shear  at  a  section  distant  x 
from  the  left  support  is 


and  the  bending  moment  at  the  same  section  is 
Mx  =  FIX  -  2Q  W(x  -a)- 


If  the  beam  is  a  cantilever,  that  is,  a  beam  fixed  in  position 
at  the  right  end  and  unsupported  at  the  left,  the  same  equation 
will  apply  when  PI  becomes  zero.  It  is  seen  by  comparing  the 
equations  above  that  F  is  always  the  first  derivative  of  M  ,  or 


Hence,  according  to  the  rule  for  determining  maxima  and  minima, 
the  bending  moment  is  always  a  maximum  (or  minimum)  at  the 
place  where  the  shear  is  zero  or  changes  in  sign.  This  criterion 
is  easily  applied  to  locate  the  points  of  M  maximum.  Pass 
along  the  beam  from  the  left  (or  right)  until  as  much  load  is  on 
the  left  (or  right)  of  the  section  as  will  neutralize  P\  (or  P%)  and 
the  point  of  M  max.  is  found.  Its  value  can  then  be  computed. 
If  the  weight  at  a  certain  point  is  more  than  enough  to  reduce 
F  to  zero,  F  changes  sign  in  passing  that  point,  and  hence  M 
max.  occurs  there. 

For  a  beam  fixed  at  one  end  only,  F  changes  sign  in  passing 
PI,  and  hence  M  max.  is  found  at  the  wall. 

Examples.  —  M  max.  occurs  in  Fig.  3,  at  C;  in  Fig.  4,  at  A;  in 
Fig.  6,  at  17.6  ft.  from  A;  in  Fig.  7,  at  A,  and  again  at  E;  in  Fig.  8, 
at  B,  and  again  at  a  distance  x  from  E  such  that  62^-  %x=  1,265!;  .'  .  x 
=\/4°-5  =  6.36  ft.;  in  Fig.  9,  at  D;  and  in  Fig.  10,  at  D  and  again  at  E. 
The  bending  moments  which  may  not  have  been  found  at  some  of 
these  points  can  now  be  computed. 

The  reader  who  is  familiar  with  graphics  can  draw  the  equi- 
librium or  bending-moment  polygons  or  curves,  and  the  shear 
diagrams,  and  notice  the  same  relation  in  them. 


46  STRUCTURAL  MECHANICS. 

The  unit  load  may  also  be  considered  as  the  derivative  of 
the  shear;  F  therefore  has  maximum  (or  minimum)  values  where 
the  external  forces  change  in  sign. 

The  origin  of  coordinates  may  be  arbitrarily  taken  at  any 
point  in  the  length  of  the  beam  and  general  expressions  may 
be  written.  If  —  iv  is  the  unit  load  and  is  constant, 


=      Fxdx  = 


in  which  FQ  and  MQ  are  the  constants  of  integration,  the  values  of 
F  and  M  at  the  origin.  Thus  in  the  beam  above,  the  bending 
moment  at  a  second  section  distant  c  from  the  first  is 

Mx+c=Mx+Fxc-Ixx+cW(x-a+c)-±wc2. 

The  same  expression  is  easily  derived  by  substituting  x+c 
for  x  in  the  original  equation. 

Example.— In  Fig.  7,  M  at  D=  +  8g6|  ft.-lb.  F  between  D  and  E= 
+  215!  Ib.  M  at  E= +8961+2151X2=1,328$  ft.-lb.  as  in  §53. 
Jfmidway  between  E  and  F=896J+  2151X3!— 300X1^=1, 2013^ 
ft.-lb. 

57.  Working  Formulas. — The  bending  moments  and  shears 
for  a  number  of  simple  cases  of  common  occurrence  are  giyen 
below.  In  general  the  bending  moment  and  the  shear  vary  from 
point  to  point  along  a  beam,  and  they  may  be  conveniently  repre- 
sented on  a  diagram  by  ordinates  whose  lengths  represent  the 
values  of  those  quantities.  In  the  accompanying  figures  the 
upper  diagram  is  one  of  bending  moments  and  the  lower  is  one 
of  shears.  Positive  values  are  laid  off  above  the  base  line  and 
negative  below.  The  student  who  has  followed  the  examples 
of  the  preceding  sections  should  have  no  difficulty  in  computing 
the  ordinates  given  in  Figs.  12  to  15. 

Figs.  12  and  13  represent  cantilever  beams,  one  carrying  a 
load  of  W  at  the  end,  the  other  carrying  a  uniformly  distributed 


BEAMS. 


47 


load  of  w  per  unit  of  length.  The  bending  moment  at  any 
section  distant  x  from  the  free  end  of  the  beam  of  Fig.  13 
is  M  =  —wx'%x  =  —  %wx2  and  the  bending-moment  diagram  is 
therefore  a  parabola. 


• 


I 

g 

! 

^         7 

Fig.  12 


Fig.  14 


Fig.  13 


Fig.  15 


Fig.  14  shows  a  beam  on  two  supports  carrying  a  load  W. 
The  bending  moment  is  evidently  a  maximum  under  the  weight, 

and  is  equal  to  W — -, — ,  a  quantity  easily  remembered  as  the 

weight  into  the  product  of  the  two  segments  divided  by  the  span. 
When  the  load  is  at  mid-span  M  =  ^WL  In  Fig.  15  a  load  of 
intensity  w  per  unit  of  length  is  distributed  over  the  beam.  The 
left  reaction  is  %wl,  and  the  bending  moment  at  a  section  dis- 
tant x  from  the  left  support  is 

M  =  %wl  -  x — wx'  \x  =  %wx(l  —x) . 

The  right-hand  member  of  this  equation  contains  x(l— x)*  which 
is  the  product  of  two  variables  whose  sum  is  constant*  therefore 
the  bending-moment  diagram  is  a  parabola. 


STRUCTURAL  MECHANICS. 


The  above  diagrams  are  drawn  for  the  applied  loads  alone, 
that  is,  the  beams  are  considered  to  be  without  weight.  If  a 
beam  weighing  iv  per  unit  of  length  carries  a  load  W  as  in  Fig. 
12  or  14,  the  bending  moment  or  shear  at  any  section  can  be 
found  by  adding  the  ordinates  of  Figs.  12  and  13  or  of  Figs,  i^ 
and  1. 


Eig.  17 


Fig.  16 


At  any  section  distant  x  from  one  end  of  a  beam  c  n  two  sup- 
ports, the  bending  moment  due  to  a  single  moving  load  is  a 
maximum  when  the  load  is  at  the  section,  when  it  has  a  value  of 
M—Wx(l-x)-^L  As  this  is  equation  of  a  parabola  of  altitude 
\Wl,  the  absolute  maximum  moments  which  can  occur  in  the  beam 
under  a  single  moving  load  are  as  shown  in  Fig.  16.  The  greatest 
positive  shear  at  any  section  occurs  when  the  load  is  just  to  the 
right  of  that  section ;  it  is  equal  to  the  left  reaction  and  consequently 
is  proportional  to  the  distance  of  that  section  from  the  right 
support,  hence  the  locus  of  maximum  positive  shear  is  a  straight 
line  as  shown. 

If  a  uniform  load  advances  continuously  from  the  right  end 
of  the  beam  of  Fig.  17,  the  positive  shear  at  any  section  will 
increase  until  the  load  reaches  the  section,  after  which  it  will 
decrease  as  the  load  extends  into  the  left  segment.  This  is  evident 
from  the  fact  that  any  load  placed  in  the  right  segment  causes 
positive  shear  at  the  section,  while  any  load  in  the  left  segment 
causes  negative  shear  at  the  section;  see  Fig.  14.  When  the 
rodd  extends  a  distance,  x,  from  the  right  support  the  shear  at 
me  head  of  the  load  is  F=Pi  =  $wx2+l  and  the  curve  of  max- 


BEAMS.  49 

imum  shears  is  a  parabola.  To  cause  maximum  negative  shean 
the  load  must  advance  from  the  left.  The  maximum  moment  at 
any  section  occurs  when  the  whole  span  is  loaded,  in  which  case 
the  moments  are  given  by  Fig.  15. 

58.  Position  of  Wheel  Concentrations  for  Maximum  Moment 
at  any  Given  Point. — Where  moving  loads  have  definite  magni- 
tudes and  spacings,  as  is  the  case  with  the  wheel  weights  of  a 
locomotive,  the  position  of  the  load,  on  a  beam  or  girder  supported 
at  both  ends,  to  give  maximum  bending  moment  at  any  given 


OoOOOO      OO      GO 


section  may  be  found  as  follows: — Let  the  given  section  be  C,  at 
a  distance  a  from  the  left  abutment  of  a  beam  A  B,  of  span  /, 
Fig.  18. 

Let  RI  be  the  resultant  of  all  loads  on  the  left  of  C,  and  acting 
at  a  distance  x\  from  the  left  abutment;  let  R2  be  the  resultant 
of  all  loads  to  the  right  of  C,  and  acting  at  a  distance  x2  from  the 
right  abutment.  The  reaction  PI  at  A,  due  to  R2  alone,  is 
R2x2  -v7,  and  the  bending  moment  at  C,  due  to  R2  only,  is  R2ax2  +1. 
Similarly  P2,  due  to  RI,  is  RiXi  +1,  and  the  bending  moment  at  C, 
due  to  RI  only,  is  R\(l—a)xi  -^/.  Hence  the  total  bending  moment 
at  C  is 


If  the  entire  system  of  loads  is  advanced  a  short  distance  d 
to  the  left,  the  bending  moment  at  C  becomes 


M  =K2 


j 


The  change  of  bending  moment  due  to  moving  the  loads  to  the 
left  is 

ad     v      - 
--—R 


2--—i 


, 


50  STRUCTURAL  MECHANICS. 

If  the  loads  are  moved  a  distance,  d,  to  the  right  instead  of 
to  the  left,  the  change  of  bending  moment  is 


From  the  two  values  of  M'  —  M  it  is  seen  that  the  bending 
moment  at  C  will  be  increased  by  moving  the  loads  to  the 
left  when  R2a>Ri(l-a)  and  by  moving  to  the  right  when 
Ri(l—a)>R2a.  When  Ri(l—a)=R2a  the  bending  moment  can- 
not be  increased  by  moving  either  to  the  right  or  left  and  is  there- 
fore a  maximum.  The  condition  may  be  written 

RI      R2      RI  -}-R2 
a     I  —  a          I 

Ri+a  is  the  average  load  per  unit  of  length  on  the  left  segment 
and  (Ri  +R2)  +1  is  the  average  load  on  the  span,  hence 

The  bending  moment  at  any  point  of  a  beam  carrying  a  system 
of  moving  loads  is  a  maximum  when  the  average  load  on  one 
segment  is  equal  to  the  average  load  on  the  span. 

Since,  for  maximum  bending  moment  at  any  section,  a  load 
must  be  at  that  section,  place  a  load  Wn  at  the  given  point  and 
compute  the  above  inequality,  first  considering  Wn  as  being  just 
to  the  right  and  then  just  to  the  left  of  the  section.  If  the  inequality 
changes  sign,  the  position  with  Wn  at  the  section  is  one  of  M  max. 
The  value  of  M  max.  can  be  computed  as  in  §  53.  If,  however, 
the  inequality  does  not  change  sign,  move  the  whole  system  until 
the  next  W  comes  to  the  section,  and  test  the  inequality  again. 

It  sometimes  happens  that  two  or  more  different  positions  of 
the  load  will  satisfy  the  condition  just  explained,  and,  to  determine 
the  absolute  M  max.,  each  must  be  worked  out  numerically. 
When  there  are  some  PF's  much  heavier  than  others,  M  max.  is 
likely  to  occur  under  some  one  of  them.  When  other  loads  are 
brought  on  at  the  right,  or  pass  off  at  the  left,  they  must  not  be 
overlooked. 

59.  Position  of  Wheel  Concentrations  for  Maximum  Shear 
at  any  Given  Point.  —  The  shear  at  point  C  in  the  beam  or  girder 
of  Fig.  19,  as  the  load  comes  on  at  the  right  end,  will  increase 
until  the  first  wheel  Wi  reaches  C.  When  that  wheel  passes  C, 
the  shear  at  that  point  suddenly  diminishes  by  W\,  and  then 
again  gradually  increases,  until  W2  reaches  C.  Let  R  be  the  sum 
of  all  loads  on  the  span  when  W\  is  at  C,  and  x  the  distance  from 
the  centre  of  gravity  of  the  loads  to  the  right  point  of  support 


OF   THE 

UNIVERSITY 


BEAMS. 


S1 


B.  The  shear  at  C  will  be  Pi  =  Rx+l.  If  the  train  moves  to 
the  left  a  distance  b,  the  space  between  W\  and  W2,  so  that 
W2  has  just  reached  C,  the  shear  at  C  will  be  R(x+b)  +l-Wi, 
plus  a  small  quantity  k  which  is  the  increase  in  PI,  due  to 
any  additional  loads  which  may  have  come  on  the  span  during 
this  advance  of  the  train.  The  shear  at  C  will  therefore  be 


0   o  o 


O  O      n  O 


Fig. 19 

increased  by  moving  up  W2,  if  Kb  +1  +k  >  TFi,  or  (as  &  can  often  be 
neglected)  if 


Rj>Wi 


or 


R    Wi 

7>TL' 


Hence  move  up  the  next  load  when  the  average  load  per  foot 
on  the  span  is  greater  than  the  load  on  the  left  divided  by  the 
distance  between  W\  and  W2. 

r>/  -  ~Dt      TT/" 

Similarly,  W%  should  be  moved  to  C  if  —r->W2  or  —  > — ?, 

If  I,  C 

Rf  being  the  sum  of  the  loads  on  the  span  when  W2  is  at  C,  and 
c  the  distance  between  W2  and  W%. 

It  is  not  necessary  to  take  account  of  k  unless  the  two  sides 
of  the  inequality  are  nearly  equal. 

Example. — Span  60  ft.,  weights  in  units  of  1,000  Ib. 

123456789     10 
Weights  =  8     15     15     15     15       9       9       9       9      8     15 
Spacing   =8'     6'     4¥   l¥     f     5'     6'     5'    8'     8' 
To  apply  test  for  M  max.  at  15  ft.  from  left,  load  advancing  from 
right.     With  W2  at  quarter  span,  load  on  span=io4.     If  W2  is  just 

to  the  right,  *°i>-  or  I2f  >1;  if  W2  is  just  to  the  left,  ^>^; 

DO         1541  4^ 

therefore  move  up  W%  to  right  of  quarter  span,     W\Q  now  is  on  the 


$2  STRUCTURAL  MECHANICS. 

span.    —  >^.     Consider  W3  to  be  just  to  the  left;  then  —  <—  , 
41  4        i 

or  the  inequality  changes  with  W%. 

PI  =  (8  -59+  15  -51  +  45  -40^+36  -2i  +  8  -5)^-60=  64.26. 
M  max.  =  (64.26)  1  5  —  8-14—  15-6=761,900  ft.-lb. 

To  test  for  F  max.  at  same  point.     Put  W\  at  quarter  span.     Load 
on  span  =95.    P>Q.    Move  up  W2]  load  on  span  now  104.    —  <-^. 

OO       o  OO          0 


Inequality    changes.     Pi  =  ( 

F.  max.  =  P  —  W  =     .2  —  S  =      2oo  Ib. 


When  these  locomotive  wheel  loads  are  distributed  to  the 
panel-joints  of  a  bridge  truss  through  the  longitudinal  stringers, 
which  span  the  panel  distance  between  floor-beams,  the  above 
rule  is  modified.  A  load  in  a  panel  being  supported  directly 
by  the  stringers  is  by  that  means  carried  to  the  joints  of  the  truss. 
When  the  train  advances  from  the  right  end  until  the  forward 
wheels  are  in  the  panel  under  investigation,  the  shear  in  that  panel 
is  the  left  reaction  minus  such  part  of  the  loads  in  the  panel  as 
go  to  the  floor-beam  to  the  left.  Let  R  be  the  resultant  of  all 
the  loads  on  the  span  applied  at  a  distance  x  from  the  right  sup- 
port, and  let  R\  be  the  resultant  of  the  loads  in  the  panel,  of  length 
/>,  applied  at  a  distance,  x\,  from  the  right  floor-beam.  Then 


- 

1    p 

If  the  loads  are  moved  a  distance,  d,  to  the  left,  the  change  of 
shear  is 


and  by  an  argument  similar  to  that  of  the  preceding  section,  for 
a  maximum 

R     Ri 


Hence  the  shear  in  any  panel  of  a  truss  is  a  maximum  when 
the  average  load  in  the  panel  is  equal  to  the  average  load  on  the 
span. 

60.  Absolute  Maximum  Bending  Moment  on  a  Beam  under 
Moving  Loads.  —  When  a  beam  or  girder  of  uniform  cross-section, 
such  as  a  rolled  I  beam,  supported  at  its  ends,  is  subjected  to 


BEAMS.  53 

a  system  of  passing  loads,  such  as  an  engine,  heavy  truck,  or 
trolley,  it  generally  suffices  to  determine  that  position  of  the 
system  of  weights  which  causes  the  absolute  maximum  bending 
moment,  the  section  where  it  is  found,  and  its  amount. 

In  Fig.  1  8  let  C  be  that  section.  Let  R  =  resultant  of  all 
loads  on  the  beam  and  #  =  its  distance  from  B;  ^1=  resultant 
of  the  loads  to  the  left  of  C.  The  reaction  at  left,  Pi  =  Rx+l', 
and,  since  the  bending  moment  at  C  is  to  be  a  maximum,  the 
shear  at  C  must  be  zero,  or 


j-i-.        ..7  =  -. 

But  the  position  of  loads  must  also  satisfy  the  condition  of 
§  58,  since  there  is  to  be  maximum  bending  moment  at  C,  and 

RI    R  RI    RI 

—  =—  .        .*.  —  =  —  ,        or        a  =  x. 

a      I  ax 

The  point  of  absolute  maximum  bending  moment  therefore 
is  as  far  from  one  end  as  the  centre  of  gravity  of  the  whole  load 
is  from  the  other.  The  rule  may  be  written  :  Place  the  loads  so 
that  the  centre  of  the  span  bisects  the  distance  between  the  centre  oj 
gravity  oj  the  whole  load  on  the  span  and  a  neighboring  wheel. 
If  the  shear  changes  sign  under  that  wheel,  the  loads  on  the  span 
are  placed  to  cause  the  greatest  possible  bending  moment. 

Example.  —  Beam  of  span  24  ft.  Two  wheels  6  ft.  apart,,  one 
carrying  2,000  Ib.  and  one  4,000  lb.,  pass  across.  Centre  of  gravity 
is  2,000  -6-f-6,ooo=  2  ft.  from  the  heavier  wheel.  Then  this  wheel  is 
to  be  placed  i  ft.  from  mid-span.  Reaction  =6,000-  11^24  =2,  7  50 
lb.  M.  ^^^.  =  2,750-11  =  30,250  ft.-lb. 


61.  Total  Tension  Equals  Total  Compression.  —  If  a  beam, 
loaded  in  any  manner,  and  in  equilibrium  under  the  moments 
caused  by  the  external  forces,  is  cut  perpendicularly  across  by 
an  imaginary  plane  of  section,  while  the  right-handed  and  left- 
handed  bending  moments  already  shown  to  exist,  §  53,  continue 
to  act,  it  is  evident  that  the  left  and  right  segments  of  the  beam 
can  only  be  restrained  from  revolving  about  this  section  by  the 
internal  stresses  exerted  between  the  material  particles  con- 
tiguous to  the  section.  These  stresses  must  be  of  such  signs, 
that  is  tensile  and  compressive;  of  such  magnitude,  provided 


54  STRUCTURAL  MECHANICS. 

the  material  does  not  give  way;  and  so  distributed  over  the 
cross-section,  as  to  make  a  resisting  moment  just  equal  to  the 
bending  moment  at  the  section.  For  the  former  is  caused  by 
the  latter  and  balances  it. 

Since  the  moment  arms  of  these  stresses  lie  in  the  perpen- 
dicular plane  of  section,  the  components  to  be  considered  now 
will  be  normal  to  the  section.  The  tangential  components  are 
caused  by  and  balance  the  external  shear. 

As  the  external  forces  which  tend  to  bend  a  beam  are  all 
transverse  to  it,  and  have  no  horizontal  components,  the  internal 
stresses  of  tension  and  compression  which  are  caused  by  the 
bending  moment  must  be  equal  and  opposite,  as  required  for 
a  moment  or  couple,  and  hence  the  total  normal  internal  tension 
on  any  section  must  equal  the  total  normal  compression. 

When  any  oblique  or  longitudinal  external  forces  act  on 
a  beam,  there  is  always  found  that  resultant  normal  stress  on 
any  right  section  which  is  required  to  give  equilibrium. 

62.  Distribution  of  Internal  Stress  on  any  Cross-section. — 
It  may  be  convenient  in  the  beginning  to  consider  one  segment 
of  the  beam  removed,  and  equilibrium  to  be  assured  between 
the  external  moment  tending  to  rotate  the  remaining  segment 
and  the  resisting  moment  developed  in  the  beam  at  the  section, 
as  shown  in  Fig.  20. 

If  two  parallel  lines  near  together  are  drawn  on  the  side  of 
a  beam,  perpendicularly  to  its  length,  before  it  is  loaded,  these 
lines,  when  the  beam  is  loaded  to  any  reasonable  amount  and 
bent  by  that  loading,  will  still  be  straight,  as  far  as  can  be 
observed  from  most  careful  examination;  but  they  will  now  con- 
verge to  a  point  known  as  the  centre  of  curvature  for  that  part 
of  the  beam. 

An  assumption,  then,  that  any  and  all  right  sections  of  the 
beam,  being  plane  before  flexure,  are  still  plane  after  the  flexure 
of  this  beam,  is  reasonable.  If  the  right  sections  become  warped, 
that  warping  would  apparently  cause  a  cumulative  endwise 
movement  of  the  particles  at  successive  sections,  especially  in  a 
beam  subjected  to  a  constant  maximum  bending  moment  over  a 
considerable  portion  of  its  span;  and  such  a  movement  and 


BEAMS. 


53 


resulting  distortion  of  the  trace  of  the  sectional  plane  ought  there- 
fore to  become  apparent  to  the  eye.  Such  a  warping  can  be 
perceived  in  shafts,  other  than  cylindrical,  subjected  to  a  twisting 
couple,  but  cannot  be  found  in  beams. 

The  lines  A  C  and  B  D  just  referred  to  will  be  found  to  be 
farther  apart  at  the  convex  side  of  the  beam,  and  nearer  together 
at  the  concave  side,  than  they  first  were;  hence  a  line  G  H,  lying 
somewhere  between  A  B  and  C  D,  is  unchanged  in  length.  If, 
in  Fig.  20,  a  line  parallel  to  A  C  is  drawn  through  H,  the  extremity 


ft    c        E  °    Fig.  20 


of  the  fibre  G  H  which  has  not  changed  in  length,  K  L,  will 
represent  the  shortening  which  I  L  has  undergone  in  its  reduction 
to  I  K,  and  N  O  will  represent  the  lengthening  which  M  N  has 
experienced  in  stretching  to  M  O.  The  lengthening  or  shorten- 
ing of  the  fibres,  whose  length  was  originally  G  TL  =  ds,  is  directly 
proportional  to  the  distance  of  the  fibre  from  G  H,  the  place  of 
no  change  of  length,  and  hence  of  no  longitudinal  or  normal 
stress. 

The  diagram,  Fig.  i,  representing  the  elongation  or  shortening 
of  a  bar  under  increasing  stresses,  shows  that,  for  stresses  within 
the  elastic  limit,  equal  increments  of  lengthening  and  shortening 
are  occasioned  by  equal  increments  of  stress.  If  this  beam  has 
not  been  loaded  so  heavily  as  to  produce  a  unit  stress  on  any 
particle  in  excess  of  the  elastic  limit  (and  no  working  beam,  one 
expected  to  last  permanently,  should  be  loaded  to  excess),  the 
longitudinal  unit  stresses  between  the  particles  will  vary  as  the 
lengthening  and  shortening  of  these  fibres,  that  is,  as  the  dis- 
tance from  the  point  of  no  stress.  Hence,  at  any  section,  the 
direct  stress  is  uniformly  varying,  with  a  maximum  tension  on 


56  STRUCTURAL  MECHANICS. 

the  convex  side  and   a  maximum  compression  on   the   concave 
side. 

The  stresses  on  different  forms  of  cross-section  A  C  are  shown 
in  Fig.  21.  The  total  tension  on  the  section  is  always  equal  to 
the  total  compression. 


63.  Neutral  Axis. — The  arrows  in  Figs.  20,  21  may  be  taken 
to  represent  the  unit  stress  at  each  point  of  the  cross-section, 
varying  as  the  distance  from  the  plane  of  no  stress,  and  constant 
in  the  direction  z.  To  locate  the  point  or  plane  of  no  stress  or 
neutral  axis  for  successive  sections : 

Let  }c  and  }t  be  the  unit  stresses  of  compression  and  tension 
between  the  particles  at  the  extreme  edge  of  any  section,  distant 
yc  and  yt  from  the  point  of  no  stress.  It  is  plain  that  }c'}t==yc:yt 
from  similar  triangles,  and  that  the  unit  stress  p  at  any  point 
distant  y  from  the  point  of  no  stress  will  be 


or 


— y,        or,  in  general  — y, 


from  a  similar  proportion. 

If  zdy  is  the  area  of  the  strip  on  which  the  unit  stress  p  is 
exerted,  z  being  the  variable  coordinate  at  right  angles  to  x  and 

y,  the  total  jorce  on  zdy  will  be  pzdy  = — yzdy,  where  —  is  a  con- 
stant, the  unit  stress  at  a  unit  distance. 

As  the  total  normal  tension  on  the  section  is  to  equal  the 
total  compression,  or  their  sum  is  to  be  zero,  §  61,  the  condition 
may  be  written 


/  r tjrye 

-  I       yzdy  =  o. 

y\K/    —yt 


Therefore  the  sum  of  the  moments  zdy-y  of  the  strips  zdy  about 
the  axis  of  z  must  balance  or  be  zero.     Then  the  axis  of  z  or 


BEAMS.  57 

neutral  axis  must  pass  through  the  centre  0}  gravity  of  a  thin  plate 
representing  the  section,  and  the  neutral  axis  of  any  section  lies 
in  its  plane,  and  usually  in  a  direction  perpendicular  to  the  plane 
oj  the  applied  external  jorces.  The  axes  of  the  successive  cross- 
sections  make  up  what  is  known  as  the  neutral  plane  of  the  beam. 
Although  there  is  no  longitudinal  or  normal  tension  or  compres- 
sion at  that  line  of  the  cross-section,  it  experiences  shear,  as  will 
be  shown  later. 

64.  Resisting  Moment.  —  The  law  of  the  variation  of  stress 
on  the  cross-section  and  the  location  of  the  neutral  axis  have  been 
established.  The  resisting  moment  is  caused  by  and  is  equal  to 
the  bending  moment.  The  moments  of  all  the  stresses  about  the 
neutral  axis  Z  Z  is,  since  p  has  the  same  sign  as  y,  and  the  moments 
conspire, 

* 


As  f+yi  denotes  the  unit  stress  at  either  extreme  fibre  divided 
by  its  distance  from  the  neutral  axis,  and  p=—yj 


where  /  represents  J  y2zdy  about  the  axis  Z  Z,  lying  in  the  plane 
of  the  section,  through  the  centre  of  gravity  of  the  same  and 
perpendicular  to  the  plane  of  the  external  forces  applied  to  the 
beam.  /  is  termed  in  mechanics  the  moment  of  inertia  of  a  plane 
area,  and  is  usually  one  of  the  principal  moments  of  inertia  of 
the  area.  The  integral  will  be  of  the  fourth  power,  involving 
the  breadth  and  the  cube  of  the  depth.  For  moments  of  inertia 
of  plane  sections,  see  Chap.  IV. 

In  the  above  expression  for  the  resisting  moment  the  quantity 
I  -z-yi  is  known  as  the  section  modulus.  The  section  moduli  of 
steel  beams,  angles,  etc.,  are  tabulated  in  the  handbooks  published 
by  the  various  steel  manufacturers,  so  that  the  resisting  moment 
of  a  steel  beam  can  be  readily  found  by  multiplying  the  section 
modulus  by  the  working  stress. 


58  STRUCTURAL  MECHANICS. 

As  moments  of  inertia  of  plane  areas  are  of  the  fourth  power, 
and  can  be  represented  by  n'bh3,  where  h  is  the  extreme  dimension 
parallel  to  y}  and  b  to  z,  and  as  y\  may  be  written  m'h,  the  resisting 
moment  can  be  represented,  if  nf  -^m'  =  n,  by 

M  =  —  =  nfbh2, 

yi 

n  being  a  fraction.     For  a  rectangular  section  this  becomes 

M=j.W^Lh*im, 

12        2  6' 

and  for  a  circular  section 

M  =  }-7^  +  I-d  =  —}d*  =  0.0982^. 

'     64         2  32' 

Examples.  —  A  timber  beam  6"Xi2//,  set  on  edge,  with  a  safe  unit 
stress  of  800  Ib.  will  safely  resist  a  bending  moment  amounting  to 
800-6-  122  -^6=  115,200  in.-lb. 

A  round  shaft,  3  in.  in  diameter,  if  /=  12,000  Ib.  will  have  a  safe 
resisting  moment  of  i2,ooo-22-33-^7'32  =  3i,82o  in.-lb. 

For  rectangular  sections,  either  b  or  h  is  usually  assumed  and 
h  or  b  then  found.  If  the  ratio  h+l  is  fixed  by  the  desire  to 
secure  a  certain  degree  of  stiffness  (see  "Deflection  of  Beams," 
Chap.  VI.),  the  unknown  quantity  is  b. 

Example.  —  A  wooden  beam  of  12  ft.  span  carries  3,600  Ib.  uniformly 
distributed.  M=%Wl=\-  3,600-  12-12  =  64,800  in.-lb.  If  /=i,ooo, 
£=1,400,000,  and  the  deflection  v  is  not  to  exceed  ^-J-j-  of  the  span, 

from  v-=--  is  obtained  -L-  -i.1-000'2-1**;  ,.  *=I3  in.    Then 

600    48  i  ,400,000  •  h 


assuming  ^=14  in.,  a  practicable  size,  64,800=-^  —  b-if;    and  b 


2   n. 


Economy  of  material  apparently  calls  for  as  large  a  value  of 
h  as  possible;  but  the  breadth  b  must  be  sufficient  to  give  lateral 
stiffness  to  the  beam,  or  it  may  fail  by  the  buckling  or  sidewise 
flexure  of  the  compression  edge,  between  those  points  where  it 


BEAMS.  59 

is  stayed  laterally.    The   effect  of    loading  as    a  beam  a  thin 
board  set  on  edge  will  make  clear  the  tendency. 

When  the  plane  of  the  applied  forces  does  not  pass  through 
the  axis  of  the  beam,  a  twisting  or  torsional  moment  is  added, 
which  will  be  discussed  in  §  86. 

65.  Limit  of  Application  of  M=f  I-^yi. — The  expression  for 
the  resisting  moment  at  any  section  of  a  beam,  caused  by  and 
always  equal  to  the  external  bending  moment  at  that  section,  is 
applicable  only  when  the  maximum  unit  stress  /  does  not  exceed 
the  unit  stress  at  the  elastic  limit  of  the  material.     If  /  exceeds 
that  limit,  a  uniformly  varying  stress  over  the  whole  section  is 
not  found,  and  the  neutral  axis  may  not  remain  at  the  centre 
of  gravity.     Hence,  also,  the  substitution  of  breaking  weights, 
obtained   by  experiments   on   beams   which  fail,  in   a  bending- 
moment  formula  which  is  then  equated  with  jl+yi,  results  in 
values  of  /,  the  then  so-called  modulus  of  rupture,  agreeing  with 
neither  the  tensile  nor  the  compressive  strength  of  the  material, 
and  therefore  of  but  limited  value.     This  formula  is  correct  for 
the  purpose  of  design  and  construction;   but  its  limitation  should 
be  kept  in  mind. 

66.  The   Smaller  Value    of   f^-yi  to  be  Used. — Since  from 
similar  triangles  }c+yc  =  lt+yt,  it  is  immaterial  which  ratio  is 
used  for  M  for  a  given  cross-section.     But,  in  designing  a  cross- 
section  to  resist  a  given  moment,  if  yt  and  yc  are  not  to  be  equal, 
another   consideration   has   weight.     A   numerical   example   will 
bring  out  the  distinction. 

A  beam  of  24  in.  span  is  loaded  at  the  middle  with  a  weight 
of  500  Ib.  M  max.  will  be  JPF/  =  5oo-6  =  3,ooo  in.-lb.  If  the 
depth  of  the  beam  is  5  in.,  and  its  section  is  of  such  a  form  that 
the  distance  from  its  centre  of  gravity  to  the  lower  edge  is  2  in., 
and  to  the  upper  edge  is  3  in.,  while  7  =  4,  then  3,000  =  ^-4 
or  J/tf«4.  Hence  the  maximum  unit  tension  /<  =  1,500  Ib.  per  sq.  in., 
and  the  maximum  unit  compression  fe=  2,250  Ib.  per  sq.  in. 
But  if  the  material  of  the  above  beam  must  not  be  subjected 
to  a  unit  stress  greater  than  2,000  Ib.  per  sq.  in.,  that  unit  stress 
will  be  found  on  the  compression  side;  for  2,000  Ib.  per  sq.  in. 
on  the  tension  side  would  be  accompanied  by  3,000  Ib.  per  sq. 


60  STRUCTURAL  MECHANICS, 

in.  on  the  compression  side;  and  a  unit  stress  of  2,000  Ib.  com- 
pression is  only  compatible  in  this  case  with  2,000-1  =  1,333  Ik- 
unit  stress  tension.  The  beam  will  safely  carry  only  a  moment 
of  2,000-4^3  =  2,667  in.-lb. 

Hence,  when  designing,  with  a  maximum  allowed  value  of 
/,  and  using  a  form  of  section  where  yt  and  yc  differ,  take  that 
ratio  of  j+yi  which  is  the  smaller.  For  a  few  materials,  where 
fe  and  ft  may  be  taken  as  differing  in  magnitude,  as  perhaps  in 
cast  iron,  use  that  ratio  }c  +ye  or  ft  +yt  which  gives  the  smaller 
value.  As  the  elastic  limit  in  tension  and  compression  for  a 
given  material  is  usually  the  same,  use  in  computations  the  larger 
value  of  yi. 

67.  Curved    Beams. — An    originally    curved    beam,    at    any 
given  cross-section  made  at  right  angles  to  its  neutral  axis,  so 
far  as  the  resisting  stresses  to  bending  moments  are  concerned, 
is  in  the  same  condition  with  an  originally  straight  beam  at  a 
similar  and  equal  cross-section  to  which  the  same  bending  moment 
is  applied.     Any  definite  thrust  or  tension  at  its  two  ends  adds 
a  moment  at  each  right  section  equal  to  the  product  of  the  force 
into  the  perpendicular  ordinate  from  the  chord  to  the  centre  of 
the  section,  and  a  force,  parallel  to  the  chord,  which  force  can 
be  resolved  into  one  normal  to  the  section  and  a  shear.     Compare 
Fig.  ii. 

68.  Inclined  Beams. — A  sloping  beam  is  to  be  treated  like 
a  horizontal  beam,  so  far  as  resisting  stress  produced  by  that 
component  of  the  load  which  is  normal  to  the  beam  is  concerned. 
The  component  of  the  load  which  acts  along  the  beam  is  to  be 
considered  as  producing  a  direct  thrust  along  the  beam  if  taken 
up  at  the  lower  end;  or  a  direct  tension,  if  taken  up  at  the  upper 
end,  or  as  divided  somewhat  indeterminately,  if  resisted  at  both 
ends.     If  this  longitudinal  force  is  axial,  the  mean  unit  stress  fe 
caused  by  it  is  to  be  added  to  the  stress  /&  of  the  same  kind  from 
bending  moment  at  the  section  where  this  sum  fe+fb  will  be  a 
maximum.     This    point    can    easily    be    found    graphically.     If 
the  section  of  the  piece  is  the  unknown  quantity,  it  will  commonly 
suffice  to  use  the  value  of  M  max.  to  determine  an  approximation 
to  /&,  and  to  correct  the  section  by  the  resulting  value  of  fe+fb 
at  the  point  where  the  sum  is  largest. 

If  the  direct  force  at  the  end  or  ends  is  not  applied  axially, 


BEAMS.  6  1 

its  moment  at  any  section  may  augment  or  diminish  the  bending 
moment  of  the  normal  components  of  the  load. 

Cases  of  inclined  beams,  for  a  given  load  and  inclination,  are 
better  solved  directly  than  by  the  application  of  formulas. 

Example.  —  A  wooden  rafter,  15  ft.  long,  has  a  horizontal  pro- 
jection of  12  ft.,  and  a  rise  of  9  ft.,  and  it  carries  a  uniformly  distributed 
load  of  1,500  Ib.  The  normal  component  of  this  load  will  be  1,200  lb., 
the  component  along  the  roof  900  lb.  The  maximum  bending  moment, 

1,200X1^X12  . 

at   the   middle,   will   be    —   —  r-^  --  =  27,000   m.-lb.     If   the   sate 

o 

T  j  2 

stress  is  1,000  lb.,  the  section  to  carry  this  moment  should  be  —  —  -  —  •  = 

27,000,  or  bh?=i62.  If  £  =  3,  h=S  in.  If  the  mean  thrust,  at  the 
middle  of  the  rafter,  is  1,250  lb.,  the  maximum  thrust,  at  the  bottom 
end,  will  be  1,700  lb.,  and  the  minimum  thrust,  at  the  top  end,  will 
be  800  lb.  The  section  of  maximum  fibre  stress  will  be  a  very  little 
below  the  middle.  But,  if  the  rafter  is  3"X8",  /&  from  bending 


moment   will   be    -~r   ,-  =  844   lb.     Also,  /C=-L-^-  =52   lb.     Hence 
3°  o-o  24 

lc+ib=&96  lb.,  a  satisfactory  result,  if  the  rafter  is  stayed  laterally  by 
the  roof-covering  or  otherwise. 

69.  Movement  of  Neutral  Axis  if  Yield-point  is  Exceeded.  — 

If  it  is  assumed  that  cross-sections  of  a  beam  still  remain  plane 
after  the  yield-point  is  passed  at  the  extreme  fibres,  the  stretch 
and  shortening  of  the  fibres  at  any  cross-section  will  continue 
to  vary  with  the  distance  from  the  neutral  axis  or  plane.  Suppose 
then  that  the  elongation  per  unit  of  length  of  the  outer  tension 
fibre  has  attained  an  amount  equal  to  O  L,  Fig.  i.  The  unit 
stress  on  that  fibre  will  be  L  N.  A  fibre  lying  half-way  from  that 
edge  to  the  neutral  axis  will  have  a  unit  stress  K  M.  If  the  beam 
is  rectangular,  the  total  tension  on  the  cross-section  must  be  the 
area  O  M  N  L,  O  L  now  being  the  distance  from  the  neutral  axis 
of  the  beam  to  the  tension  edge.  Since  the  total  compression  on 
the  section  must  equal  the  total  tension,  an  equal  area  O  L'N' 
must  be  cut  off  by  L'  N'  and  the  compression  curve.  The  neu- 
tral axis  must  then  divide  the  given  depth  of  the  beam  in  the 
ratio  of  O  L  to  O  L',  shifting  in  this  case  towards  the  compression 
side.  Had  the  compression  curve  been  below  the  tension  curve, 
the  neutral  axis  would  have  shifted  towards  the  convex  side  of 
the  beam. 

Since  L  N  is  less  than  L'  N',  the  unit  stress  on  the  extreme 
fibre  on  the  tension  side  is  the  less.  Hence  this  displacement  of 
the  neutral  axis  favors  the  weaker  side.  If  such  action  continued 


62  STRUCTURAL  MECHANICS. 

to  the  time  of  fracture,  it  would  account  for  the  fact  that  the 
application  of  the  usual  formula,  jl  +y\,  to  breaking  moments 
gives  a  value  of  /  which  lies  between  the  ultimate  tensile  and 
compressive  strengths  of  the  material.  It  must  be  borne  in  mind, 
however,  that  the  compression  portion  of  the  section  increases 
in  breadth  and  the  tension  portion  contracts,  quite  materially 
for  ductile  substances,  thus  adding  to  the  complication.  A 
soft  steel  bar  cannot  be  broken  by  flexure  as  a  beam  at  a  single 
test. 

A  rectangular  cross-section  also  tends  to  assume  the  section 
shown  in  Fig.  22.  The  compressed  particles  in  the  middle  of  the 
width  can  move  up  more  readily  than  they  can 
laterally,  making  the  upper  surface  convex  as  well 
as  wider,  and  the  particles  below  at  the  edges, 
being  drawn  or  forced  in,  are  crowded  down, 
making  the  lower  surface  concave  as  well  as 
narrower. 

Hence  the  position  of  the  neutral  axis  is  un- 
certain, after  the  yield-point  has  been  passed  on 
either  face;  but  it  is  probably  moved  towards  the  stronger  side. 

70.  Cross-section  of  Equal  Strength.  —  When  a  material  will 
safely  resist  greater  compression  than  tension,  or  the  reverse, 
it  is  sometimes  the  custom  to  use  such  a  form  of  cross-section 
that  the  centre  of  gravity  lies  nearer  the  weaker  side.  Cast 
iron  is  properly  used  in  sections  of  this  sort.  See  Fig.  21,  section 
at  right.  Wrought-iron  or  steel  sections  are  occasionally  rolled 
or  built  up  in  a  similar  fashion,  but  the  increase  in  width  of  the 
compression  flange  is  then  usually  intended  to  increase  its  lateral 
stiffness. 

If  /j=safe  unit  tensile  stress,  and  /c=safe  unit  compressive 
stress,  the  centre  of  gravity  of  the  section  must  be  found  at 
such  point  that  yt'yc"=}t'}ci  when  the  given  safe  stresses  will 
occur  simultaneously  at  the  section.  By  composition,  yt'-Jc'-h 
=  }t'}c'-}t  +}c>  so  that  the  centre  of  gravity  should  be  distant 
from  the  bottom  or  top, 


Example.  —  If  ^=3,000  lb.,  and  /c=9,ooo  lb.,  ^=^-3,000-^-12,000 
\h.  If  a  cast-iron  X  section  is  to  be  used,  base  10  in.,  thickness  through- 


BEAMS.  63 

out  of  i  in.,  and  height  of  web  hf,  then,  by  moments  around  base, 

,. 
=ij  in.; 

8  in,lb., 


12      12  4  4'3 

the  moment  that  the  section  will  carry. 

71.  Beam  of  Uniform  Strength.  —  As  has  been  shown  in 
§  64,  the  resisting  moment  may  be  put  into  the  form  M  =  njbh2, 
where  n  is  a  numerical  factor  depending  on  the  form  of  cross  - 
section.  If,  then,  for  a  given  load,  bh2  be  varied  at  successive 
cross-sections  to  correspond  with  the  variation  of  the  external 
bending  moment,  the  unit  stress  on  the  extreme  fibre  will  be 
constant;  the  beam  will  be  equally  strong  at  all  sections,  except 
against  shear;  and  there  will  be  no  waste  of  material  for  a  given 
type  of  cross-section,  provided  material  is  not  wasted  in  shaping. 

Suppose,  for  example,  that  a  beam  is  to  be  supported  at 
its  ends,  to  carry  W  at  the  middle,  and  to  be  rectangular  in  cross  - 
section.  The  bending  moment  at  any  point  between  one  sup- 
port and  the  middle  is  %Wx.  Equate  this  value  with  the  resist- 
ing moment.  ^Wx=^fbh2.  To  make  /  constant  at  all  cross- 
sections,  bh2  must  vary  as  x  from  each  end  to  the  middle.  If  h 
is  constant,  b  must  vary  as  xt  or  the  beam  will  be  lozenge-shaped 
in  plan  and  rectangular  in  elevation.  If,  on  the  other  hand, 
b  is  constant,  h2  must  vary  as  xt  and  the  elevation  will  consist 
of  two  parabolas  with  vertices  at  the  ends  of  the  beam  and  axis 
horizontal,  while  the  plan  will  be  rectangular. 

The  section  need  not  be  a  rectangle.  If  the  ratio  of  b  to 
h  is  not  fixed,  the  treatment  will  be  like  the  above;  but,  if  that 
ratio  is  fixed,  as  for  a  circular  section,  or  other  regular  figure, 
b  =  ch,  and  h3  must  vary  as  the  external  bending  moment,  or, 
in  the  case  above,  as  x.  The  cross-section  of  the  cast-iron  beam 
in  the  example  of  the  previous  section  may  be  varied  in  accordance 
with  these  principles. 

The  following  table  gives  the  shape  of  beams  of  rectangular 
cross-section  supported  and  loaded  as  stated. 

When  a  beam  supported  at  both  ends  carries  a  single  moving 
load  W,  passing  across  the  beam,  the  bending  moment  at  the 


64 


STRUCTURAL  MECHANICS. 


Beam. 

M. 

bh2 
varies  as 

h2  Constant, 
b  varies  as 

b  Constant, 
h2  varies  as 

Fixed  at  one  end, 

-Wx 

X 

x,  triangular  plan, 

xy  parabolic  eleva- 

W at  other. 

Fig.  23. 

tion.     Fig.  24. 

Fixed  at  one  end, 

—  $wx* 

X2 

x2,  parabolic  plan, 

x2,  h  varies  as  x, 

uniform  load. 

Fig.  25. 

triangular  eleva- 

r 

tion.     Fig.  26. 

SupYdbothends  [ 
W  at  a  from  ] 
end. 

W—x 
Wa 

X 

l-x 

!  triangular  plan, 
/_aJ       Fig'  27> 

x      1  parabolic  ele- 
>     vation. 
/_J      Fig.  28. 

Sup't'd  both  ends, 

^wx(l-x} 

x(l-x) 

x(l—  x)  parabolic  plan. 

x(l—  x},      circular 

uniform  load. 

Fig.  29. 

or  elliptical  eleva- 

tion.    Fig.  30. 

point  x,  where  the  load  is  at  any  instant,  —Wx(l—x)  -t-l.     Such 
a  beam  will  therefore  fall  under  the  last  class  of  the  above  table. 


Fig.  23 


Fig.  27 


Fig.  24 


Fig.  25 


Fig-  28 


Fig.  29 


Fig.  26 


Fig.  30 


BEAMS.  65 

Beams  which  can  be  cast  in  form  or  built  up  may  be  made 
in  the  above  outlines,  if  desired.  Some  common  examples, 
such  as  brackets,  girders  of  varying  depth,  walking-beams,  cranks, 
grate-bars,  etc.,  are  more  or  less  close  approximations  to  such 
forms.  Enough  material  must  also  be  found  at  any  section 
to  resist  the  shear,  as  at  the  ends  of  beams  supported  at  the  ends. 

Where  a  plate  girder  is  used  (see  Fig.  95)  with  a  constant 
depth,  the  cross -section  of  the  flanges,  or  their  thickness  when 
their  breadth  is  constant,  will  theoretically  and  approximately 
follow  the  fourth  column  of  the  preceding  table.  If  the  flange 
section  is  to  be  constant  or  nearly  so,  the  depth  must  vary  in  the 
same  way,  and  not  as  in  the  fifth  column. 

Roof-  and  bridge-trusses  are  beams  of  approximate  uniform 
strength,  for  the  different  allowable  unit  stresses  and  for  chang- 
ing loads.  The  principles  of  this  section  have  an  influence  on 
the  choice  of  outline  for  such  trusses,  and  the  shapes  of  moment 
diagrams  suggest  truss  forms. 

72.  Distribution  of  Shearing  Stress  in  the  Section  of  a  Beam, 
Pin,  etc. — It  will  be  proved,  in  §  154,  that,  at  any  point  in  a 
body  under  stress,  the  unit  shear  on  a  pair  of  planes  at  right 
angles  must  be  equal.  Whatever  can  be  proved  true  in  regard 
to  the  unit  shear  on  a  longitudinal  plane  at  any  point  in  a  beam 
must  therefore  be  true  of  the  unit  shear  on  a  transverse  plane  at 
the  same  point. 

Fig.  31  represents  a  portion  of  a  beam  bent  under  any  load. 
The  existence  of  shear  on  planes  parallel  to  E  F  is  shown  by  the 
tendency  of  the  layers  to  slide  by  one 
another  upon  flexure.  Let  the  cross - 
section  of  the  beam  be  constant.  If 
the  bending  moment  at  section  H,  a 
point  close  to  G,  differs  from  that  at  G, 
there  will  be  a  shear  on  the  transverse 

section,  because  the  shear  is  the  first  derivative  of  the  bending 
moment,  §  56.  The  direct  stress,  here  compression,  on  the 
face  H  F  of  the  solid  H  F  E  G,  will  differ  from  that  on  the  face 
G  E,  since  the  bending  moments  are  different,  and  that  difference 
will  be  balanced  by  a  longitudinal  horizontal  force,  or  shear, 


66  STRUCTURAL  MECHANICS. 

on  the  plane  FE,  to  oppose  the  tendency  to  displacement.  If 
this  force  along  the  plane  E  F  is  divided  by  the  area  E  F  over 
which  it  is  distributed,  the  longitudinal  unit  shear  will  be  obtained. 
It  follows  from  the  first  paragraph  that  the  unit  shear  at  the 
point  E  on  the  transverse  section  G  A  must  be  the  same.  It  is 
also  evident  that  the  farther  E  F  is  taken  from  H  G,  the  greater 
will  be  the  difference  between  the  total  force  on  H  F  and  that 
on  G  E,  until  the  neutral  axis  is  reached,  and  that  the  unit  shear 
on  the  longitudinal  plane  E  F  must  increase  as  E  F  approaches 
B,  the  neutral  axis.  The  same  thing  is  true  if  the  plane  is  sup- 
posed to  lie  at  different  distances  from  the  edge  A.  Hence,  at 
any  transverse  section  A  G,  the  unit  shear  on  a  longitudinal  plane 
is  most  intense  at  the  neutral  axis  ;  and  therefore  the  unit  shear 
on  a  transverse  section  A  G  is  unequally  distributed,  being  greatest 
at  B,  the  neutral  axis,  and  diminishing  to  zero  at  A  and  G. 

Pins  and  keys,  and  rivets  which  do  not  fit  tightly  in  their 
holes,  and  hence  are  exposed  to  bending,  have  a  maximum  unit 
shear  at  the  centre  of  any  cross-section,  and  this  shear  must 
therefore  be  greater  than  the  mean  value,  and  must  determine 
the  necessary  section. 

To  find  the  mathematical  expression  for  the  variation  of 
shear  on  the  plane  A  G: 

O  B  D  C  is  the  trace  of  the  neutral  plane.  B  D  =  E  F  sensibly 
=  dx.  B  E=y,  B  G  =  ^I.  Breadth  of  beam  at  any  point  =  2,  at 
neutral  axis  =  z0.  Normal  or  direct  unit  stress  at  the  point  E 
on  plane  A  G  =  p.  Unit  shear  at  E  =  ^;  maximum,  at  B,  =  q$. 
M  and  F=  bending  moment  and  shearing  force  at  section  A  G. 


By  §  64,  }=--    and    p=-j-y* 

The  total  direct  stress  on  plane  G  E  is 

/vi  M  fvi 

P*dy=j-Jy    yzdy..    .....    (i) 

The  difference  between  M  at  the  section  through  B  and  M 
at  the  section  through  D  must  be  Fdx,  since  M  =  jFdx,  by  §  56. 


BEAMS.  67 

The  horizontal  force  on  E  F  is  the  excess  of  (i)  for  G  E  over 

Fdx  Cy^ 
its  value  for  H  F,  or  —j—  I      yzdy.     Divide  by  the    area    zydx 

1     J  y 

of  F  E,  over  which  this  horizontal  force  acts,  to  find  the  unit  shear. 

F    f*  F    fyi 

^=7T/    yzdy'    ?0=77/n    yzdy' 

IZyJ  y  IZQt/O 

Since  the  mean  unit  shear  =  F+  S,  the  ratio  of  the  maximum 
unit  shear  to  the  mean  will  be  found  by  dividing  qo  by  F+S. 


ryi 

S     fyi  Jo 

Mean  unit  shear    Iz0Jo    y    y         r2z0 


Max,  unit  shear 


where  r=  radius   of  gyration   of   the  cross-section,   and    fyzdy 

is  the  moment  of  either  the  upper  or  lower  part  of  the  cross- 
section  about  the  trace  of  the  neutral  plane.  Hence  the  max- 
imum unit  shear  will  be: 

r& 

b  I      ydy  79 

Jo     J  J      12  -h2      * 
Rectangle,          2          =~='  or  5°%  greater- 


^?  °r  33  greater- 

Thin  ring  approximately  =  2,  or  100  per  cent,  greater  than  the 
mean  unit  shear. 

For  beams  of  variable  cross  -section  /  will  not  be  constant; 
but  the  preceding  results  are  near  enough  the  truth  for  prac' 
tical  purposes. 

Example.  —  A  cylindrical  bridge-pin  3  in.  diam.,  area  7.07  sq.  in., 

has  a  shear  of  50,000  Ib.     The  maximum  unit  shear  is  —  -  .A=» 

7.07     3 
9,430  Ib.  per  sq.  in. 

If  the  apparent  allowable  unit  shear  is  reduced  one  quarter, 
ao  from  10,000  Ib.  to  7,500  Ib.,  the  same  circular  section  for  a 


$8  STRUCTURAL  MECHANICS. 

pin  will  be  obtained  in  designing  as  if  the  maximum  unit  shear 
were  considered.  For  a  rectangular  section  the  apparent  allow- 
able shear  should  be  reduced  one-third. 

Example.  —  A  4"X6"  beam  has  at  a  certain  section  a  shear  of  2,400 
Ib.  ;  the  maximum  unit  shear  on  both  the  horizontal  and  the  vertical 

plane,  at  the  middle  of  the  depth,  is  ----  =150  Ib.  per  sq.  in. 

As  the  shearing  resistance  along  the  grain  of  timber  is  much 
less  than  the  shearing  resistance  across  the  grain,  wooden  beams 
which  fail  by  shearing  fracture  along  the  grain  at  or  near  the 
neutral  axis,  at  that  section  where  the  external  shear  is  greatest. 
As  the  unit  shears  on  two  planes  at  right  angles  through  a  given 
point  are  always  equal,  the  shearing  strength  of  timber  across 
the  grain  cannot  be  availed  of,  since  the  piece  will  always  shear 
along  the  grain. 

A  rectangular  timber  beam  of  span  /,  carrying  w  per  unit 
of  length,  has  a  maximum  fibre  stress  of 


and  the  greatest  shearing  stress  along  the  neutral  axis  is 


Dividing  the  first  equation  by  the  second  gives  the  ratio  between 
the  maximum  fibre  stress  and  the  maximum  shear  existing  in 
the  beam.  If  the  corresponding  ratio  between  the  allowable 
stresses  for  the  beam  is  greater  than  the  ratio  between  the  existing 
stresses,  the  beam  is  weaker  in  shear  than  in  flexure;  hence  if  / 
and  q0  are  working  unit  stresses,  a  rectangular  beam  carrying  a 
uniformly  distributed  load  should  be  designed  for  shear  when 


BEAMS.  69 

A  beam  carrying  a  single  load  at  the  centre  should  be  designed 

for  shear  when 


h  -2#o 

73.  Variation    of   Unit  Shear. — The   distribution    of   shear 
on  three  forms  of  cross-section  is  indicated  in  Fig.  32,  where 


Jig.  32 

the  ordinates  show  the  unit  shear  at  corresponding  points.  For 
both  the  rectangle  and  the  circle  the  intensity  of  shear  varies 
according  to  the  ordinates  of  a  parabola.  The  curve  for  the 
I-shaped  section  is  made  up  of  three  parabolas  as  shown,  the 
intensity  of  shear  being  nearly  constant  over  the  web.  For 
I  beams  of  ordinary  proportions  and  for  plate  girders  the  curve 
showing  unit  shears  in  the  web  will  be  much  flatter  than  in  this 
figure,  and  the  maximum  unit  shear  will  differ  but  little  from 
the  unit  shear  found  by  dividing  the  total  shear  at  a  section  by 
Mie  cross-section  of  the  web,  as  is  usually  done  in  practice. 

Examples. — i.  Three  men  carry  a  uniform  timber  30  ft.  long. 
One  man  holds  one  end  of  the  timber;  the  other  two  support  the 
beam  on  a  handspike  between  them.  Place  the  handspike  so  that 
each  of  the  two  shall  carry  f  of  the  weight. 

2.  Three  sections  of  water-pipe,  each  12  ft.  long,  are  leaded  end 
to  end.     In  lowering  them  into  the  trench,  where  shall  the  two  slings 
be  placed  so  that  the  joints  will  not  be  strained?     Neglect  the  extra 
"Weight  of  socket. 

3.  Wooden  floor-joists  of  14  ft.  span  and  spaced  12  in.  from  centre 
to  centre  are  expected  to  carry  a  floor  load  of  80  Ib.  per  sq.  ft.     If 
7=900  Ib.,  what  is  a  suitable  size?  2//Xio//. 

4.  One  of  these  joists  comes  at  the  side  of  an  opening  4'X6',  the 
load  from  the  shorter  joists,  then  10  ft.  long,  being  brought  on  this 
longer  joist  at  4  ft.  from  one  end.     How  thick  should  this  joist  be  ? 

4  in. 


70  STRUCTURAL  MECHANICS. 

5.  A  cylindrical  water-tank,  radius  20  ft.,  is  supported  on  I  beams 
radiating  from  the  centre.     These  beams  are  supported  at  one  end 
under  the  centre  of  the  tank  and  also  on  a  circular  girder  of  15  ft. 
radius.     They  are  spaced  3  ft.  apart  at  their  outer  extremities.     If 
the  load  is  2,000  Ib.  per  sq.  ft.  of  bottom  of  tank,  find  the  max.+  and 
— Jkfonabeam.  +29,600;   —68,750  ft.-lb. 

6.  Find  b  and  h  for  the  strongest  rectangular  beam  that  can  be 
sawed  from  a  round  log  of  diameter  d. 

7.  A  opening  10  ft.  wide,  in  a  i6-in.  brick  wall,  is  spanned  by  a 
beam  supported  at  its  ends.     The  maximum  load  will  be  a  triangle 
of  brick  3  ft.  high  at  the  mid-span.     If  the  brickwork  weighs  112  Ib. 
per  cu.  ft.,  find  M  at  the  mid-span.  44,800  in.-lb. 

8.  In  the  above  problem,  write  an  expression  for  M  at  any  point 
if  w= weight  of  unit  volume  of  load,  a= height  of  load  at  middle, 
/= thickness  of  wall,  /=span,  and  x= distance  of  point  from  the  sup- 
port. 

9.  A  trolley  weighing  2,000  Ib.  runs  across  a  beam  6  in.  wide  and 
of  20  ft.  span.     What  will  be  the  elevation  of  a  beam  of  uniform  strength, 
and  what  its  depth  at  middle,  if  /=8oo  Ib.  ?  12  in.+ 

10.  A  round  steel  pin  is  acted  upon  by  two  forces  perpendicular 
to  its  axis,  a  thrust  of  3,000  Ib.  applied  at  8  in.  from  the  fixed  end  of 
the  pin,  and  a  pull  of  2,000  Ib.  applied  6  in.  from  the  fixed  end  and 
making  an  angle  of  60°  with  the  direction  of  the  first  force.     Find 
the  size  of  the  pin,  if  /=  8,000  Ib.  M=  20,784  in.-lb. 

11.  A  beam  of  20  ft.  span  carries  two  wheels  6  ft.  apart  longitudi- 
nally, and  weighing  8,000  Ib.  each.     When  they  pass  across  the  span, 
where  and  what  is  M  max.  ?  57>8oo  ft.-lb. 

12.  A  floor-beam  for  a  bridge  spans  the  roadway  a  and  projects 
under  each  sidewalk  b.     If  dead  load  per  foot  is  w,  live  load  for  road- 
way ix/}  for  sidewalk  w",  write  expressions  for  +M  max.  and  —  M  max. 

13.  A  vessel  is  200  ft.  long.     It  carries  5  tons  per  ft.  uniformly 
distributed,  and  a  central  load  of  300  tons.     Find  M  max.  when  at 
rest;    when  supported  on  a  wave  crest  at  bow  and  stern  with  each 
bearing  20  ft.  long;  and  when  supported  amidships  only  with  bearing 
30  ft.  long. 

14.  The  end  of  a  beam  6  in.  wide  is  built  into  a  wall  18  in.     The 
bending  moment  at  the  wall  is  600,000  in.-lb.     If  the  top  of  the  beam 
bears  for  9  in.  with  a  uniformly  varying  pressure  and  the  bottom  the 
same,  what  is  the  maximum  unit  compression  on  the  bearing  surface  ? 

1,852  Ib. 


CHAPTER  IV. 
MOMENTS  OF  INERTIA  OF  PLANE  AREAS, 

74.  Definitions.  —  The  rectangular  moment  of  inertia,  7,  of 
a  plane  area  about  an  axis  lying  in  that  plane  is  the  sum  of  the 
products  of  each  elementary  area  into  the  square  of  its  distance 
from  the  axis.  The  plane  areas  whose  moments  of  inertia  are 
sought  are  commonly  cross  -sections  of  beams,  and  unless  other- 
wise stated  the  axis  about  which  moments  are  taken  passes 
through  the  centre  of  gravity  of  the  cross-section.  The  quotient 
of  the  moment  of  inertia  by  the  area  is  the  square  of  the  radius 
oj  gyration,  r.  If  the  area  be  referred  to  rectangular  axes  Z 
and  Y,  and  if  subscripts  denote  the  axes  about  which  moments 
are  taken, 

/,  =  ffyVtdy  =  fzy*dy  =  Sr.»  ; 


If  each  elementary  area  be  multiplied  by  the  square  of  its 
distance  from  an  axis  perpendicular  to  the  plane  and  passing 
through  the  centre  of  gravity,  the  summation  gives  the  polar 
moment  of  inertia,  J.  As  the  distance  of  each  elementary  area 
from  the  axis  is  Vz2  +y2, 


When  the  term  "moment  of  inertia"  is  used  without  qualification, 
the  rectangular  moment  is  meant.  Moments  of  inertia,  being 
the  product  of  an  area  into  the  square  of  a  distance,  are  of  the 
fourth  power  and  positive. 


7  a  STRUCTURAL  MECHANICS. 

The  values  of  7,  7,  and  r2  for  some  common  forms  of  cross- 
section  follow. 

I.  Rectangle,  height  h,  base  6.  Fig.  33.  Axis  through  the 
centre  of  gravity  and  parallel  to  b. 


,.  +  &  /•  +  **  j-  -,+tf       ^3       ^3       ^3 

,=   /        y2zdy  =  b  y2dy=\-by3\       =  —  +  —  =—  . 

J-&   '  J-ih  '  l_3       J-^     24      24      12 


12  12  12 

For  an  axis  through  the  centre  of  gravity  and  perpendicular 
to  the  plane, 

bh  T 

y         Z       12  12 

II.  Triangle,  height  h,  base  b.     Fig.  34.     Axis  as  above  and 
parallel  to  b. 

2    7  ^    /2     7 

h:b  =  —  h—y:z'}        z=j-[—h- 
3  'l  \3 

/+ffc              b     r  +  %h/2          \             b  ["2            i     "1** 
y2zdy  =  -r   I         I  ~/i  —  y  \y2dy  =  7-1  — /z,^3 ^4  I 

16       16        2         i  \         &/£3 


III.  Isosceles    triangle,    about    axis    of   symmetry.     Fig.    35. 
Height  along  axis  h,  base  b. 


Ie  —  2/~     '"V-1-         A  //  *"/  ——-|    0        „/,!..         -*'V«^       oo  /        48 


2  A 


MOMENTS  OF   INERTIA   OF  PLANE  AREAS. 


73 


The  sum  of  II  and  III  will  be  the  polar  moment,  7,  about 
an  axis  through  the  centre  of  gravity  and  perpendicular  to  the 
plane. 


123 


*3 


Fig.  33 


Fig.  34 


Fig.  35 


Fig.  36 


Fig.  37 


IV.  Circle,  radius  R,  diameter  d.  Fig.  36.  If  0  =  angle 
between  the  axis  of  Z  and  a  radius  drawn  to  the  extremity  of  any 
element  parallel  to  that  axis, 


=  R  sin  d; 


/,= 


0;        dy=RcosOdd. 

i  fi  1?     *R4 

•ii  sin  40-0      =  — 
8L.4  4 


7i  R2     4          1 6 

The  polar  moment  of  inertia,  /,  may  be  easily  written  if 
r'  =  variable  radius, 


Since  Iz+Iy  =  J,  and  Iy  =  Ig  by  symmetry,  Iz  =  ^7tR4  as  before. 

V.  Ellipse.     Diameters  d  and  6.     Fig.  37. 

As  the  value  of  z  in  the  ellipse  is  to  that  of  z  in  the  circle, 
as  the  respective  horizontal  diameters,  or  as  b  to  d,  and  as  the 
moment  of  the  strip  zdy  varies  as  the  breadth  alone,  the  ellipse 
having  horizontal  diameter  6,  height  d,  gives 


64 


64 


16' 


ndlfi 


74  STRUCTURAL  MECHANICS. 

VI.  The  moment  of  inertia  of  a  hollow  section,  when  the 
areas  bounded  respectively  by  the  exterior  and  interior  perimeters 
have  a  common  axis  through  their  centres  of  gravity,  can  be 
found  by  subtracting  /  for  the  latter  from  /  for  the  former.  Thus  : 

Hollow  rectangle,  interior  dimensions  b'  and  hf,  exterior  b 
and/*;  Iz  =  ^(bh*-bfh'*}. 

Hollow  circle,  interior  radius  R',  exterior  radius  R; 


The  moment  of  inertia  of  a  hollow  ring  of  outside  diameter 
d  and  inside  diameter  d'  ',  the  ratio  of  dr  to  d  being  n,  may  be 
written 


When  the  thickness  /  of  a  ring  is  small  as  compared  with  the 
diameter, 

Ig^jxtd3  nearly. 

75.  Moment  of  Inertia  about  a  Parallel  Axis.  —  To  find  the 
moment  of  inertia  /'  of  a  plane  area  about  an  axis  Z  parallel 
to  the  axis  Z0  through  the  centre  of  gravity  and  distant  c  from  it. 

By    definition    I'  =  f  (y  +c)2zdy  =  j  y2zdy  +  2CJ  yzdy  +c2j  zdy. 

The  first  term  of  the  second  member  is  Iz,  the  moment  of  inertia 
about  the  axis  through  the  centre  of  gravity;  the  second  term 
has  for  its  integral  the  moment  of  the  area  about  its  centre  of 
gravity,  which  moment  is  zero  ;  and  the  integral  in  the  third  term 
is  the  given  area  S.  Hence 

/'  =  Iz  +c2S  =  (r2  +c2)S  =  S2S. 

Example.  —  J2  for  rectangle,  axis  parallel  to  b,  is  J?bh3.    7'  about 
^=4^3,    and  r'2=$h2. 


The  reverse  process  is  convenient  for  use. 
Iz  =  If  -c2S  =  (r'2  -c2)S  -  r2S. 


MOMENTS  OF  INERTIA   OF  PLANE  AREAS.  75 

As  the  value  of  /  about  an  axis  through  the  centre  of  gravity 
is  the  least  of  all  7's  about  parallel  axes,  it  can  readily  be  seen 
whether  c2S  is  to  be  added  or  subtracted. 

If  the  value  of  /i  about  an  axis  distant  Ci  fr6m  the  centre 
of  gravity  is  known,  and  it  is  desired  to  find  1  2  about  a  parallel 
axis  distant  c2  from  the  centre  of  gravity,  a  combination  of  the 
two  formulas 


and 
gves 


Example.  —  I  for  a  triangle  about  an  axis  through  the  vertex  parallel 
to  the  base  is  easily  obtained,  since  z:b=y:h. 

/hb  bh3 

—y3dy=  —  . 
h,  4 

_,  ^3/i      4\j2bh     bit* 

Then  /  about  base  =  ---  \-  [  --  -   h2—  =  —  . 

4       \9      9/      2       12 

76.  Moments  of  Inertia  of  Shapes.  —  It  is  frequently  necessary 
to  divide  areas,  such  as  T,  I,  and  built  iron  sections,  and  those 
of  irregular  outline,  into  parts  whose  moments  of  inertia  are 
known,  each  about  an  axis  through  its  own  centre  of  gravity; 
then  to  the  sum  of  their  several  7's  add  the  sum  of  the  products 
of  each  smaller  area  into  the  square  of  the  distance  from  its 
axis  to  the  parallel  axis  through  the  centre  of  gravity  of  the  whole. 
This  rule  is  an  expansion  of  the  preceding  one. 

Iz  for  the  whole  =  21  +  Ic2S. 

Formulas  for  such  cases  are  of  little  value.  In  actual  com- 
putation follow  the  general  rule. 

Example.  —  Find  /  of  a  6"X4"Xi"  angle  about  an  axis  parallel  to 
shorter  leg.  Neglecting  fillets  as  is  customary,  S=  9.  5X^=4.75  sq.  in. 
Divide  figure  into  two  parts,  6X^=3  sq.  in.  and  3.5X^=1.75  sq.  in. 
Distance  of  centre  of  gravity  from  middle  of  6-in.  leg  is  3.  5X0.5  X  2.75-7- 
4.75=1.01  in.,  which  is  1.99  in.  from  heel. 


Or  consider  two  rectangles,  one  4"X6",  the  other  3i"X5i". 
/=TVX4X63+4X6Xi.oi2— T^X  3.5X5. 53— 3. 5X5. 5X1. 252=17.40  in. 


76 


STRUCTURAL   MECHANICS. 


77.  Moments  of  Inertia  for  Thin  Sections. — Values  of  7  for 
rolled  shapes  may  also  be  approximately  obtained  by  the  follow- 
ing method,  and,  if  the  values  given  in  the  manufacturers'  hand- 
books are  not  at  hand,  will  prove  serviceable. 

The  moment   of  inertia   of   a   thin  strip   or 
y\   rod,  Fig.  38,  of  length  L  and  thickness  /,  about 

-8. 1  an  axis  passing  through  one  end  of  and  making 

an  angle  6  with  it,  is  the  same  as  if  J/Z,  were 
concentrated  at  the  extreme  end.  Let  /  =  distance  along  strip 
to  any  particle. 


•  fL 

Jo 


tdl  •  I2  sin2  0 


sin2  0  ? 


or  one-third  the  area  multiplied  by  the  square  of  the  ordinate 
to  the  extreme  end. 

This  expression  might  be  derived  from  /  for  a  rectangle, 
taken  about  one  base. 

If  the  rod  is  parallel  to  the  axis,  and  at  a  distance  y±  from 
it,  I  =  tL-yi2,  since  all  particles  are  equidistant  from  the  axis. 

Example.  —  Find  approximate  value  of  /  of  angle  of  last  example. 
Using  centre  line,  angle  is  5J"X3i"Xj".  Distance  of  centre  of 
gravity  from  heel  is  5.75X2.875-1-9.5  =  1.74  in. 


78.  Rotation  of  Axes.  —  If  the  moments  of  inertia  about 
two  axes  at  right  angles  are  known,  the  moments  of  inertia  about 
axes  making  an  angle  a  with 
the  first  may  be  found.  In 

the  solution  the 


or    j  I  zydzdy  occurs,    which  is 

called    the    product    of    inertia 
and    represented    by    Z.    Prod- 
ucts  of    inertia    may   be   either 
positive    or    negative    according 
to  the  position  of  the  axes.     Fiom  Fig.  39, 
zf  =  z  cos  a  +y  sin  a, 
y  =  y  cos  a—  z  sin  OL\ 


MOMENTS  OF  INERTIA   OF  PLANE  AREAS.  77 

7/  =  j  (y  cos  a  —  z  sin  a)2dS 

=  Jy2  cos2  adS  —  2jzy  sin  a  cos  adS  +Jz2  sin2  adS, 
Iy  =  I  (z  cos  a+y  sin  a)2dS 

=  f  z2  cos2  adS  -4-2  /  z-y  sin  a  cos  a<£S  +  I  y2  sin2  a</5 
«/  •/  •/ 

Z'  —  \  (z  cos  a:  +y  sin  a) (;y  cos  a:  —2  sin  a)(ZS 

=  J(y2  —z2)  sin  a  cos  a^/5  +J  zy  cos2  adS—j zy  sin2  adS; 

7/  =  7g  cos2  a+Iy  sin2  a—2Z sin  a  cos  a, 
7y'  =  Iy  cos2  oL+Ig  sin2  a  +  2Z  sin  a:  cos  a, 

Zr  =  (7 z  —Iy)  sin  a  cos  a  +Z  (cos2  a  —sin2  a). 

79.  Principal  Axes. — To  find  the  maximum  and  minimum 
values  of  //  and  Iy  differentiate  with  respect  to  a : 

—r^-=  —2lz  sin  a  cos  a  +2ly  sin  a  cos  a  —  2Z  cos2  a.  +2Z  sin2  a, 


Hence  maximum  and  minimum  values  of  //  and  //  occur  when 
Z'  =  o,  and  since  J  =  Iz+Iy  =  //+//  =  a  constant,  //  is  a  mini- 
mum when  //  is  a  maximum.  The  axes  for  which  //  and  Ivf 
are  maximum  and  minimum  are  called  principal  axes.  An 
axis  of  symmetry  is  always  a  principal  axis  since  for  such  an 
axis  Z  must  be  zero,  as  for  every  positive  ordinate  there  is  a  cor- 
responding negative  ordinate.  If  a  figure  has  two  axes  of  sym- 
metry not  at  right  angles,  the  moment  of  inertia  is  the  same  for 
all  axes  through  the  centre  of  gravity. 

The  angle,  <£,  which  the  principal  axis,  A  A,  makes  with  the 
original  Z  axis  may  be  found  by  making  Z'=o: 

(Iz  -Iy}  sin  <£  cos  $  +Z  (cos2  <f>  -sin2  <£)  =o, 
5—  Iy)  sin  2<£ -HZ  cos  2^=0, 
2Z 


STRUCTURAL  MECHANICS. 


IA,  the  moment  of  inertia  about  the  A  axis,  is  found  by  substi- 
tuting the  value  of  <j>  in  the  expression  for  I  /  : 


IA  =  Iz  cos2  0  +/i/  sin2  0  —Z  sin  20  ; 

los2  0  =  K1  +cos  2<£)>         sm2  <t>  =  4C1  "~cos 

^  =i/s(i  +cos  20)  +  J/j,(i  —cos  20)  —  Z  sin 

^J(/2/-/2)(i  -cos  20)  -Z  sin  20-f-/r 


tan  20 

cos  20 -i         t 

—  ^      •       i      i  •**  > 

sin  20 

IA=IZ—  Z  tan  0. 


tan  0  = 


I-COS20> 

sin  20    ' 


Similarly  the  moment  of  inertia  about  axis    B   B  at  right 
angles  to  axis  A  A  is  found  to  be 


tan  0. 


80.  Oblique  Loading.  —  When  the  plane  of  loading  on  a 
beam  does  not  coincide  with  one  of  the  principal  axes  of  the 
section  the  beam  does  not  deflect  in  the  plane  of  the  loads  and 
the  neutral  axis  is  oblique  to  that  plane.  In  a  cross-section  of 
such  a  beam  the  stress  at  a  point  whose  coordinates  are  a  and 
b  referred  to  the  A  and  B  axes  may  be  found  by  resolving  the 


bending  moment  at  the  section  into  its  components  in  the  direction 
of  each  principal  axis,  finding  the  stress  at  the  point  due  to  each 


MOMENTS  OF  INERTIA   OF  PLANE  AREAS.  79 

component   and   adding  algebraically.     Thus   the  stress  at  the 
point  D,  Fig.  40,  due  to  a  bending  moment,  M,  is 

(Msm6)b     (M  cos  6)  a 

=    ~  ~ 


It  may  sometimes  be  necessary  to  find  the  direction  of  the 
neutral  axis  to  determine  which  is  the  extreme  or  most  stressed 
fibre.  Since  there  is  no  stress  at  the  neutral  axis,  if  the  last 
equation  be  made  equal  to  zero,  a  and  b  will  become  coordinates 
of  some  point  on  the  neutral  axis,  and  the  ratio  between  them 
will  be  the  tangent  of  the  angle,  /?,  which  the  neutral  axis  makes 
with  the  A  axis, 

tan/3  =  -  =  -T^ctnfl. 
a        IB 

Example.  —  A  6X4Xi-in.  angle  is  used  as  a  purlin  on  a  roof  of 
}  pitch  (tan"1^!),  as  shown  in  Fig.  40.  What  is  the  maximum  fibre 
stress  due  to  a  bending  moment  M?  Centre  of  gravity  lies  0.99  in. 
from  back  of  longer  leg  and  1.99  in.  from  shorter. 

72=17.40  in.4,        7^=6.27  in.4. 


=  +6.08  in4. 

tan  20=  2X6.08-^(6.27—17.40)  =  —  1.092, 
0=  —  23°  46',        tan  <f>=  —0.440. 

IA=  17.40—  6.  o8(—  0.440)=  20.07,       7^  =  6.  27  +  6.  o8(—  0.440)  =  3.  60. 
6=  tan"1  1.5  —  0=8o°  04',     51110=0.985,     cos#=o.i73,     ctn#=o.i75. 
tan/?=  —  20.07X0.175-^3.60=  —0.978,        ft—  —44°  02'. 

Laying  off  angles  and  scaling  gives 

a=i.98in.,       £=3.05  in. 


.o5  ~2o.o>j)  +  M  (0.173X1.  98  -^  3.  60)  =  0.243  If. 
If  the  load  is  applied  along  the  Y  axis, 

0  =  9o°  -0=ii30  46', 

sin  6=0.  915,        cos  6=—  0.403,       ctn  #=  —  0.440. 
tan  p=  20.  07X0.  440-^3.  60=  2.  460,          /?=67°  53'. 


So  STRUCTURAL  MECHANICS. 

The  greatest  stress  is  found  on  the  inside  edge  of  the  lower  leg,  where 

a=  +  i.i8in.,        b=—  3.88  in. 
/=  —M  (0.915X3.88-^  20.07)  —  Af  (0.403X1. 18-^3. 6o)=—o.^ogM. 

Examples. — i.  Find  the  moment  of  inertia  of  a  trapezoid,  bases  a 
and  b,  height  h,  about  one  base. 

2.  A  i2-in.  joist  has  two  mortises  cut  through  it,  each  2  in.  square, 
and  2  in!  from  edge  of  joist  to  edge  of  mortise.     How  much  is  that 
section  of  the  joist  weakened  ?  ^  or  26%. 

3.  A  bridge  floor  is  made  of  plates  rolled  to  half -hexagon  troughs, 
6  in.  face,  5.2  in.  deep,  12  in.  opening,  J  in.  thick.     Find  the  resisting 
moment  of  a  section  18  in.  wide.  20.8/. 

4.  If  that  floor  is  14  ft.  between  trusses  and  carries  two  rails  5  ft. 
apart,  each  loaded  with  2,000  Ib.  per  running  foot,  what  will  be  the 
unit  stress?  7>79°  ID; 

5.  Six  thin  rolled  shapes,  web  a,  make  a  hexagonal  column,  radius 
a,  with  riveted  outside  flanges,  each  b  in  width.     Prove  that 


CHAPTER  V. 
TORSION. 

8  1.  Torsional  Moment.  —  If  a  uniform  cylindrical  bar  is 
twisted  by  applying  equal  and  opposite  couples  or  moments  at 
two  points  of  the  axis,  the  planes  of  the  couples  being  perpen- 
dicular to  that  axis,  the  particles  on  one  side  of  a  cross-section 
tend  to  rotate  about  the  axis  and  past  the  particles  on  the  other 
side  of  the  section,  thus  developing  a  shearing  stress  that  varies 
with  the  tendency  to  displacement  of  the  particles,  that  is,  directly 
as  the  distance  of  each  particle  from  the  centre.  The  unit  shear 
then  is  constant  on  any  ring,  and  the  shearing  stresses  thus  set 
up  at  any  section  make  up  the  resisting  moment  to  the  torsional 
moment  of  the  applied  couple.  As  all  cross-sections  are  equal 
and  the  torsional  moment  is  constant  between  the  two  points 
first  referred  to,  each  longitudinal  fibre  will  take  the  form  of  a 
helix. 

82.  Torsional  Moment  of  a  Cylinder.  —  If  the  unit  shear  at 
the  circumference  of  the  outer  circle,  Fig.  41,  of  radius  RI  and 
diameter  d  is  q\,  the  value  at  a  distance  R  from  the  centre  will 
be,  by  the  above  statement,  q  =  qiR+Ri.  The  total  shearing 
force  on  the  face  of  an  infinitesimal  particle  whose  lever-arm  is 

R,  and  area  RdRdd,  will  be  ~R*dRdd,  and  its  moment  about 

^-i 
ql 
the  centre  will  be  —R3dRdd.     Hence  the  resisting  moment  for 


a  cylinder  is 


*        27C  ^-  =  - 

2\i         2 

81 


82  STRUCTURAL  MECHANICS. 

Hence  the  resisting  moment  against  torsion  resembles  in 
form  the  resisting  moment  against  flexure,  but  differs  in  using 
the  polar  moment  of  inertia  of  the  cross-section  for  the  rectangular 
one,  and  in  having  q\,  maximum  unit  shear,  in  place  of  /,  maxi- 
mum unit  tension  or  compression. 

As  the  rings  of  metal  situated  farthest  from  the  centre  of  a 
shaft  offer  the  greatest  resistance  to  torsion,  it  is  economical  of 
metal  to  make  the  shaft  hollow.  If  d  is  the  internal  and  D  the 
external  diameter  of  a  shaft,  the  resisting  moment  is  found  by 
inserting  the  expression  for  the  polar  moment  of  inertia  of  a  ring 
in  the  last  equation,  hence 


. 

Example.  —  Design   a   round   shaft   to   transmit    500  H.P.   at   100 
revolutions  per  minute  if  #1=8,000  Ib.  per  sq.  in. 

500X33,000-^-100=165,000  ft.-lb.  of  work  per  revolution. 

Assume  this  work  to  be  performed  by  a  force  P  rotating  about  the 
centre  of  the  shaft  at  a  distance  unity.  The  work  done  by  P  in  one 
revolution  will  be  the  force  into  the  distance  through  which  it  acts, 
or  27rP,  and  the  torque,  T,  will  be  PX  i  ;  hence 

T=i  65,000-^-2^  =  26,  300  ft.-lb.  =  315,000  in.-lb. 


0.196X8,000 


=201,         d=6  in.  nearly. 


Fig-  4* 


83.  Twist  of  a  Cylindrical  Shaft. — If  the  surface  of  a  cylin- 
drical shaft  is  divided  into  small  squares  by  two  sets  of  lines, 
one  set  running  along  the  elements  and  the  other  at  right  angles 
to  them,  the  first  set  will  become  helices  when  the  shaft  is  twisted, 
while  the  second  set  will  remain  unchanged  with  the  result  that 


TORSION.  83 

the  squares  will  become  rhombs,  as  is  shown  in  Fig.  42.  The 
angle  of  distortion  of  the  rhombs  is  <j>,  which  is  the  angle  that 
every  tangent  to  a  helix  makes  with  the  axis  of  the  shaft.  When 
the  free  end  of  the  shaft  has  rotated  by  the  fixed  end  through 
an  angle  6,  for  small  distortions  such  as  occur  in  practice  (f>l=Ri6. 
From  §  173,  (j>  =  qi+-C,  hence 


Example.  —  A  round  shaft  2^  in.  diameter  carries  a  pulley  of  30  in. 
diameter;  the  difference  in  tension  on  the  two  parts  of  the  belt  is 
1,000  ib.  Then  7^=1,000-15  =  15,000  in.-lb.  if  the  torsional  moment 
is  entirely  carried  by  the  section  of  the  shaft  on  one  side  of  the  pulley. 
If  the  shaft  is  30  ft.  long  and  C=  11,500,000, 


IO.2-  i. 
0=—  —±-          —  =0.122. 

1  1,  500,000  -54 

To  reduce  this  angle  to  degrees  multiply  by  180^-^=57.3  and  obtain 
6°  59'.     £1  =  4*890  Ib.  per  sq.  in. 

84.  Non-Cylindrical  Shafts.  —  The  formulas  deduced  for  cylin- 
drical shafts  cannot  be  applied  to  shafts  of  any  other  form  since 
sections  which  are  not  circular  become  warped  under  torsion 
and  the  unit  shear  is  not  proportional  to  its  distance  from  the 
axis  of  the  shaft.  On  any  form  of  cross-section  the  shear  at  the 
perimeter  must  be  tangential;  for,  if  it  is  not,  it  can  be  resolved 
into  two  components,  one  tangential  and  one  normal.  By  §  154 
the  normal  component  would  necessitate  an  equal  shear  on  a  plane, 
at  right  angles  ;  that  is,  along  the  surface  of  the  shaft  parallel  to 
the  axis,  which  is  impossible.  St.-Venant  deduced  theoretically 
formulas  for  prisms  of  various  cross-sections  under  torsion.  The 
solution  is  very  involved  and  only  his  results  will  be  given.  In 
each  case  the  maximum  stress  is  found  at  points  on  the  perimeter 
which  lie  nearest  the  axis.  If  the  prism  has  corners  or  edges, 
the  stress  there  is  zero.  The  notation  is  the  same  as  in  §  74. 

Rectangle,  T  =  —,  -  rtfi  ; 

15/^+96^ 

Square,  T  =  o.  208^1  h3  ; 

Ellipse,  T=o.ig6qibd2- 

Equilateral  triangle,     T=  0.07  7^  h?  =  0.050^1  63. 


STRUCTURAL  MECHANICS. 


The  twist  of  any  of  these  forms  is  very  nearly 

TJ 

e=4°c^i. 

85.  Bach's  Formula. — Professor  Bach  deduced  an  expression 
for  the  torsional  moment  of  a  rectangular  shaft  from  experiment. 
The  sides  of  a  rectangular  steel  bar  were  ruled  into  squares  and 
the  bar  was  twisted  to  give  it  a  permanent  set.  The  squares  were 
distorted  by  the  twist  into  rhombs  and  the  amount  of  the  distortion 
of  any  square  was  taken  to  be  proportional  to  the  intensity  of  the 
shear.  The  distortion  was  greatest  at  the  middle  of  the  longer  side, 
at  the  middle  of  the  shorter  side  it  bore  the  relation  of  b  to  h  to 
that  of  the  longer  side,  and  at  the  corners  it  was  zero.  The  variation 
along  one  side  was  given  very  well  by  the  ordinates  to  a  parabola 
with  the  vertex  opposite  the  middle  of  the  side.  The  principal  axes 
of  any  original  right  section  remained  straight  and  at  right  angles 
to  each  other  after  the  bar  was  twisted.  The  unit  shears  at  points 
on  these  axes  were  assumed  to  be  proportional  to  their  distances 
from  the  centre  of  the  section  and  to  act  at  right  angles  to  the  axes. 
From  these  assumptions  this  equation  was  deduced: 


Rectangle, 


T=%qlb2h. 


h>b 


86.  Combined  Bending  and  Torsion.  —  A  shaft  is  often 
subjected  to  bending  in  addition  to  torsion,  and  in  such  a  case 
must  be  designed  to  resist  both.  The  normal  unit  stress  on 
the  cross-section  at  the  extreme  fibre  from  the  bending  moment 
on  the  beam  or  shaft  is  /,  tension  on  one  edge,  compression  on 
the  opposite  edge. 

The  unit  shear  on  the  same  cross-section  at  the  same  extreme 
fibre  is  gi,  and  as  the  shears  on  two  planes  at  right  angles  are 


\                *',B          \.- 

—  ->B          \J 

w 

>/x 

MD-V 

\ 

J  -<,  - 

-A     , 

Pig,  48 


equal  by  §  154,  the  stresses  at  the  extreme  fibre  on  a  longitudinal 
and  on  a  transverse  plane  are  as  shown  in  Fig.  43.     (The  plane 


TORSION.  85 

of  the  bending  moment  is  at  right  angles  to  the  plane  of  the 
paper.)  The  principal  stresses  can  be  found  by  the  method  of 
§168. 

AB  is  the  plane  of  cross-section;  NO=/;  RN  =  <?i.  Then 
RO  is  the  resultant  unit  stress  on  AB,  R'O  =  q\  on  second  plane 
revolved  90°  to  make  the  two  planes  and  normals  coincide.  Draw 
RR'  connecting  the  extremities  of  the  two  stresses.  As  its  middle 
point  falls  on  the  middle  point,  M,  of  NO, 


By  §  152,  fe» 

Since  M=jl+y\  and  T  =  qiJ-^yi  and  since  2/=7  for  a 
square  or  a  circle,  (i)  may  be  multiplied  by  /  -±yi  and  transformed 
to 


(2) 


in  which  M  =  original  bending  moment  at  the  section,  T  =  original 
torsional  moment,  and  M  i  =  equivalent  resulting  bending  moment 
for  which  the  shaft  should  be  designed  so  that  the  unit  stress 
shall  not  exceed  /  when  both  M  and  T  occur  at  the  given  section. 
If  (i)  is  multiplied  by  J  -±-y\,  there  results 


as  an  equivalent  torsional  moment  in  which  TI=  p\J  -±y\. 
Although  neither  equation  (2)  nor  (3)  is  rational,  since  pi  does  not 
act  on  a  right  section,  the  conception  of  an  equivalent  bending 
moment  is  less  objectionable  than  the  conception  of  an  equivalent 
torsional  moment  because  p\  is  a  normal  stress.  But  in  designing 
shafting  it  is  usual  to  employ  the  same  working  value  for  unit 
fibre  stress,  /,  as  for  unit  shear,  q\\  consequently  the  results 
obtained  are  the  same  whether  equation  (i),  (2),  or  (3)  is  used. 
If  the  deformation  of  the  material  is  taken  into  account  as  in 


36  STRUCTURAL  MECHANICS. 

§  175,  the    new    principal    stresses    become    pi/=spi^^p2    and 


Examples.  —  i.  If  the  pulley  of  the  previous  example  weighs  500  Ib. 
and  is  12  in.  from  the  hanger,  on  a  free  end  of  the  shaft,  *and  the  un- 
balanced belt-pull  of  1,000  Ib.  is  horizontal,  the  resultant  force  on  the 
pulley  is  5ooVi2+22=  1,120  Ib.,  which  causes  a  bending  moment  of 
13,400  in.-lb.  at  the  hanger.  Then 


o2)}  =  16,750  in.-lb., 
which  will  cause  a  fibre  stress  of 


ovx   0=10,950  ID.  per  sq.  m. 
0.098X5* 

2.  The  wooden  roller  of  a  windlass  is  4  ft.  between  bearings.     What 
should  be  its  diameter  to  safely  lift  4,000  Ib.  with  a  2-in.  rope  and 
a  crank  at  each  end,  both  cranks  being  used  and  /  being  800  Ib.  ? 

3.  Design  a  shaft  to  transmit  500  horse-power  at  80  revolutions 
per  rain.*  if  £1=9,000  Ib.  d=6  in. 

4.  How  large  a  shaft  will  be  required  to  resist  a  torsional  moment 
of  i, 600  ft.-lb.  if  £1  =  7,500  Ib.  ?     If  the  shaft  is  75  ft.  long  and  C= 
11,200,000,  what  will  be  the  angle  of  torsion?  2f  in.;  30^°. 

5.  What  torsional  moment  will  a  hollow  shaft  of  5  in.  internal 
and  10  in.  external  diameter  transmit  if  £1  =  8,000  Ib.  per  sq.  in.? 
What  is  the  size  of  a  solid  shaft  to  transmit  the  same  torsional  moimnt  ? 
What  is  the  difference  in  weight  between  the  two  shafts  if  a  bar  of  rteel 
of  one  inch  section  and  one  foot  long  weighs  3.4  Ib.? 


CHAPTER  VI. 


FLEXURE  AND  DEFLECTION  OF  SIMPLE  BEAMS. 

87.  Introduction. — As  the  stresses  of  tension  and  compression 
which  make  up  the  resisting  moment  at  any  section  of  a  beam 
cause  elongation  and  shortening  of  the  respective  longitudinal 
elements  or  layers  on  either  side  of  the  neutral  plane,  a  curvature 
of  the  beam  will  result.  This  curvature  will  be  found  to  depend 
upon  the  material  used  for  the  beam,  upon  the  magnitude  and 
distribution  of  the  load,  the  span  of  the  beam  and  manner  of 
support,  and  upon  the  dimensions  and  form  of  cross-section. 
It  is  at  times  desirable  to  ascertain  the  amount  of  deflection,  or 
perpendicular  displacement  from  its  original  position,  of  any 
point,  or  of  the  most  displaced  point,  of  any  given  beam  carrying 
a  given  load. 

Further,  the  investigation  of  the  forces  and  moments  which  act 
on  beams  supported  in  any  other  than  the  ways  already  discussed 
requires  the  use  of  equations  that  take 
account  of  the  bending  of  the  beams 
under  these  moments.  There  are  too 
many  unknown  quantities  to  admit  of  a 
solution  by  the  principles  of  statics  alone. 
The  required  equations  involve,  expres- 
sions for  the  inclination  or  slope  of  the 
tangent  to  the  curved  neutral  axis  of 
the  bent  beam  at  any  point,  and  its 
deflection,  or  perpendicular  displacement, 
at  any  point  from  its  original  straight 
line,  or  from  a  given  axis.  The  curve 
assumed  by  the  neutral  axis  of  the  bent  beam  is  called  the  elastic 
curve. 

87 


A 


88  STRUCTURAL  MECHANICS. 

88.  Formula  for  Curvature.  —  If,  through  the  points  A  and 
B,  on  the  elastic  curve,  Fig.  44,  and  distant  ds  apart,  normals 
C  D  and  K  G  to  the  curve  of  this  neutral  axis  are  drawn,  the 
distance  from  A  B  to  their  intersection  will  be  the  radius  of 
curvature  p  for  that  portion  of  the  curve.     If  through  A  a  plane 
F  H  is  passed  parallel  to  K  G,  the  distance  F  C  will  be  the  elon- 
gation, or  H  D  will  be  the  shortening,  from  the  unit  stress  /,  of 
the  extreme   fibre  which    was    ds  long  before    flexure.     Cross- 
sections  plane  before  flexure  are  plane  after  flexure,  §  62. 

AO=|0;    AC=^i;    CF  =  pds,  §  10.     From  similar  triangles 
ACF   and   O  A  B,   p'.ds-yujds,   or   <o=^p      As,   by    §64, 

—  ,    — 

the  curvature  or  the  amount  of  bending  at  any  one  point. 

89.  Slope  and  Deflection.  —  If  the  elastic  curve  is  referred  to 
rectangular  coordinates,  x  being  measured  parallel  to  the  original 
straight  axis  of  the  beam,  and  v  perpendicular  to  the  same,  the 
calculus  gives  for  the  radius  of  curvature 


M  =  —  ,    —=—,  the  reciprocal  of  the  radius  of  curvature,  called 


fflv 

dx2 

For  very  slight  curvature,  such  as  exists  in  practical,  safe  beams, 

dv  . 

—  is  a  very  small  quantity,  and  in  comparison  with  unity  its  square 

may  be  neglected.    Then 

i d2v    M 

f)     dx2    EF 

As  M  is  a  function  of  x,  as  has  been  seen  already,  the  first 
definite  integral,  -7-,  will  give  the  tangent  of  the  inclination  or  the 
slope  of  the  tangent  to  the  curve  of  the  neutral  axis  at  any  point 


FLEXURE  AND  DEFLECTION   OF  SIMPLE  BEAMS.          89 

x,  and  the  second  integral  will  give  v,  the  deflection,  or  perpen- 
dicular ordinate  to  the  curve  from  the  axis  of  x.    Thus 

dv 

Aslope,         ,; 

di  d2v 

£7  —  =  moment,  M  = 


dM  d*v 

-j-  =  shear,  £  =  £/—; 

dx  dx3' 

dF  d*v 

—  =  load,  w—EI  3-7. 
dx 


While  the  following  applications  of  the  operations  indicated 
in  the  last  paragraph  are  examples  only,  the  results  in  most  of 
them  will  be  serviceable  for  reference. 

The  student  must  be  careful,  in  solving  problems  of  this  class, 
to  use  a  general  value  for  M,  and  not  M  maximum.  The  origin 
of  coordinates  will  be  taken  at  a  point  of  support,  such  a  point 
being  definitely  located;  x  is  measured  horizontally,  v  vertically, 
and  —  v  denotes  deflection  downwards. 

The  greatest  deflection  for  a  given  load  is  v^^.  The  greatest 
allowable  deflection  for  a  fibre  stress  /  is  v\. 

If  M+I  is  constant,  the  beam  bends  to  the  arc  of  a  circle. 
This  happens  where  M  is  constant  and  /  is  constant,  or  where 
I  varies  as  M. 

Example.  —  The  middle  segment  of  a  timber  beam  carrying  a  load, 
W,  at  each  quarter-point  bends  to  the  arc  of  a  circle.  M=  }PT/;  p=  -. 


If  the  beam  is  6  in.Xi2  in.X2o  ft.,  W=  2,000  Ib.  and  £=1,500,000 

4  •  i  ,  q  00,000  •  864 
Ib.  per  sq.  in.,  then  7=TV-6-i23=  864   in.4   and  p=  -  -  — 

2,000-20-12 

=  10,800  in.  =  900  ft. 

90.  Beam  Fixed  at  One  End;  Single  Load  at  the  other;  origin 
at  the  wall;  length  =/.  Let  the  abscissa  of  any  point  on  the 
elastic  curve  be  x,  and  the  ordinate  v.  At  that  point 

Mx=-W(l-x). 


9o 


STRUCTURAL  MECHANICS. 


Then 


dx2 


The  slope  at  the  point  x  is  found  by  integrating. 
dv         W 

• 


At  the  point  where  x  =  o,  the  slope  is  zero  and  therefore  C=o. 
Integrating  again, 

W 


When  X  =  QJ  v=o,  therefore  C' =  o.     The  equation  just  found  is 
the  equation  of  the  elastic  curve.     If  x  is  made  equal  to  /  in  the 


above  equations,  the  slope  and  the  deflection  at  the  end  of  the 
beam  are  found  to  be 

Wl2 


To  determine  the  maximum  allowable  deflection  of  a  given 
beam  consistent  with  a  safe  unit  stress  in  the  extreme  fibre  at  the 
section  of  maximum  bending  moment,  substitute,  in  the  expression 
for  vmax,,  the  value  of  W  in  terms  of  /. 


y\  yd' 

Wl3     ft2 


FLEXURE  AND   DEFLECTION   OF  SIMPLE  BEAMS. 


91 


7  = 


Example. — If /=6o  in.,  b=4  in.,  h=S>  in.,  W  = 

512  8oo-6o2-3 

- — ;      max.    slope  =  - 
3  2-1,400,000-512 


.,£=1,400,000; 

0.006;         7W*.  = 


800  •  6o3  •  3 

—  =  0.24   m. ;    and  if  /=  1,200  lb.,   maximum  safe  de- 
3-1,400,000-512 

i,2ooX6o2  ,  . 

nection=  — - —          — —  =  0.20  m. 
3X1,400,000X4 

It  will  be  seen  that,  for  a  given  weight,  the  maximum  bending 
moment  varies  as  the  length  /;  the  maximum  slope  varies  as  P, 
and  the  maximum  deflection  as  I3.  The  slope  and  deflection 
also  vary  inversely  as  7,  or  inversely  as  the  breadth  and  the  cube 
of  the  depth  of  the  beam.  The  maximum  safe  deflection,  how- 
ever, consistent  with  the  working  unit  stress  /,  varies  as  /2,  and 
inversely  as  y\,  or  the  depth  of  the  beam.  These  relationships 
are  true  for  other  cases,  as  will  be  seen  in  what  follows. 

The  ease  with  which  problems  regarding  deflection  are  solved 
depends  greatly  upon  the  point  taken  for  the  origin,  as  it  influences 
the  value  of  the  constants  of  integration. 

91.  Beam  Fixed  at  One  End;  Uniform  Load  of  w  per  unit 
over  the  whole  length,  /;  origin  at  the  wall. 


When  x=o,  * 


dx 

.;  .'.  C=o 


92  STRUCTURAL  MECHANICS. 

When  oc=o,  v=o;  .'.  C'=o.     When  x=l, 


8EI' 

For  maximum  safe  deflection  consistent  with  unit  stress,  /, 
in  the  extreme  fibre  at  the  dangerous  section, 


ft2 


SEI 


92.  Combination   of  Uniform  Load  and  Single  Load  at  one 

end  of  a  beam  fixed  at  the  other  end.     Add  the  corresponding 
values  of  the  two  cases  preceding. 


Note,  in  the  expression  for  vmax.,  the  relative  deflections 
due  to  a  load  at  the  end  and  to  the  same  load  distributed  along 
the  beam;  and  compare  with  the  respective  maximum  bending 
moments. 

Example.  —  If  the  preceding  beam  weighs   50  lb.,  the  additional 

deflection  will  be  -  --  =o.ooc;  in.,  too  small  a  quantity  to 
8-1,400,000-512 

be  of  importance.     In  the  majority  of  cases  the  weight  of  the  beam 
itself  may  be  neglected,  unless  the  span  is  long. 

93.  Beam  Supported  at  Both  Ends;  Uniform  Load  of  w 
per  unit  over  the  whole  length  /;  origin  at  left  point  of  support. 


v      w 
___ 


FLEXURE  AND  DEFLECTION   OF  SIMPLE  BEAMS 


93 


From  conditions  of  symmetry  it  is  evident  that  the  slope  is 
zero  when  #  =  |/;  hence,  to  determine  the  constant  of  integration, 
make 


As  v=o  when  #  =  o  or  /,  C'=o.     When  #= 


5    Q/)/3 


Fig.  47 


For  maximum  safe   deflection  consistent  with  a  unit  stress, 
/,  in  the  extreme  fibre  at  the  middle  section, 


* 


«-^;      "i=- 


Examples. .--A  pine  floor-joist,  uniformly  loaded,  section  2X12  in., 
span  14  ft.=  i68  in.,  has  deflected  f  in.  at  the  middle.  Is  it  safe? 
£=1,500,000.  By  the  last  formula, 


I2 


4      48    1,500,000-6' 


/=  2,300  lb., 


and  the  beam  is  overloaded. 

What  weight  is  it  carrying  ?     By  formula  for  vntax,t 


i3  •  1 23  • 


T2 


1,500,000-2-  12' 


^=5,248  lb. 


94  STRUCTURAL  MECHANICS. 

94.  Beam  Supported  at  Both  Ends ;   Single  Load  W  at  middle 
of  span  /;  origin  at  left  point  of  support. 

d2v      W 


This  expression  will  apply  only  from  #=o  to  x  =  \l\  but, 
as  the  two  halves  of  the  deflection  curve  are  symmetrical,  the 
discussion  of  the  left  half  will  suffice. 


When  *=i/,    *-o.    .'.  C=  -J/2. 

W 


When  #=o,  v=o.    .*.  C'=o.    The  limit  x=l  is  not  applicable. 


For  maximum  safe  deflection  consistent  with  a  unit  fibre 

H  AlT  W 

stress,  /, 


Example.  —  A  lo-in.  steel  I  beam  of  33  Ib.  per  ft.  and  7=  162;  span, 
15  ft.,=  i8o  in.,  carries  in  addition  a  uniform  load  of  767  Ib.  per  ft. 
of  span  and  6,000  Ib.  concentrated  at  the  middle.  What  will  be  its 
deflection  and  the  maximum  unit  fibre  stress? 


«  800-  15  ,  6,ooo\  i8o3 


FLEXURE  AND  DEFLECTION  OF  SIMPLE  BEAMS.  95 

95.  Single  Weight  on  Beam  of  Span  /,  supported  at  both 
ends;  W  at  a  given  distance  a  from  the  origin,  which  is  at  the 
left  point  of  support. 


Fig-  48 


As  the  ioaJ  L  eccentiic,  the  curve  of  the  beam  is  unsym- 
metrical,  and  equations  must  be  written  for  each  portion,  oc<a 
and  x>a. 


ON  LEFT  OF  WEIGHT. 


tfv      W 


(i) 


v  =  o,     when     x=o;     .".  C"=o. 


ON  RIGHT  OF  WEIGHT. 

Wa.  _Wa_         , 

~T~  '        Mx  —  —j-(l  —  X). 

W  . 


(2) 


v=o,  when  x  =  l; 

.'.  C'"=-ja/3-C'/. 


For  equations  to  determine  the  constants  C  and  C,  use  the 
value  x  =  a,  when  it  will  be  evident  that  i  for  the  left  segment 
must  give  the  same  value  as  does  i  for  the  right  segment,  and 
v  at  a  must  be  the  same  when  obtained  from  the  left  column  as 
when  obtained  from  the  right.  Therefore,  from  (i)  and  (2), 


v  at  a  gives 


C'n  =  ^cfil 

dx=E7/\~r*+T~6~ 3/' 
V==~ETl\~X '  ~*~~2   ~~6        3  /' 


W 


<z/2 


_W_(aW_     ax*     a*x     al2x     a*l\ 
"£7/^2         6        6"~"1~+T/ 


96  STRUCTURAL  MECHANICS. 

As  a  is  assumed  to  be  less  than  £/,  and  the  substitution  of 

dv 
x  =  a  in  the  value  of  -r-  on  the  left  gives  a  slope  which  is  negative, 

the  point  of  vmax.  will  be  found  on  the  right  of  W,  and  for  that 
value  of  x  which  makes  -7-  on  the  right  zero.     Hence 


-  alx + J<z3  +  Ja/2 = o. 


which  is  the  distance  from  the  left-hand  support  to  the  point  of 
maximum  deflection.  Substitute  in  the  expression  for  v  on  the 
right,  to  obtain  the  maximum  deflection. 

It  should  be  noticed  that,  when  the  weight  is  eccentric,  the 
point  of  maximum  deflection  is  found  between  the  weight  and 
the  mid-span,  and  not  at  the  point  of  maximum  bending  moment, 
which  latter  is  under  the  weight. 

96.  Two  Equal  Weights  on  Beam  of  Span  /,  supported  at 
the  ends  ;  each  W,  symmetrically  placed,  distant  a  from  one  end. 

This  case  may  be  solved  by  itself,  but  can  be  more  readily 
treated  by  reference  to  §  95.  Thus  the  maximum  deflection 
will  be  at  the  middle,  and  can  be  found  by  making  #  =  J/in  the 
above  value  of  v  for  the  right  segment  and  doubling  the  result. 
Then 


The  deflection  under  a  weight  will  be  given  by  the  addition 
of  v  at  a  and  v  at  (/  —  a)  of  the  preceding  case.    Thus 

Wa2  Wa2 

-a*       Vt-°= 

Wa2 


FLEXURE   AND  DEFLECTION  OF  SIMPLE  BEAMS.  97 

Example. — A  round  iron  bar,  12  ft.  long  and  2  in.  diameter,  carries 
two  weights  of  200  Ib.  each  at  points  3  ft.  distant  from  either  of  the 
two  supported  ends.  The  deflection  at  a  weight = 

200  •  3  62  •  7  •  4 

2---i2  =  o.6  in. 


6  -28,000,000-  22  w 

The  maximum  unit  bending  stress  is  — • — —  =  9,160  Ib. 


22 


DEFLECTION   OF   BEAMS   OF   UNIFORM   STRENGTH. 

It  will  be  apparent  that  a  beam  of  uniform  strength  will  not 
be  so  stiff  as  a  corresponding  beam  of  uniform  section  sufficient 
to  carry  safely  the  maximum  bending  moment;  for  the  stiffness 
arising  from  the  additional  material  in  the  second  case  is  lost. 

97.  Uniform  Strength  and  Uniform  Depth.  (See  §  71.)  — 
Since  M=njbh2  and  varies  as  bh2,  and  I=n'bh3  and  varies  as 
bh3,  M  -±1  varies  as  i+h.  But  if  h  is  constant,  M  +1  is  constant 

d2v 
and  -7-3  is  constant.    Therefore  all  beams  of  this  class  bend  to 

dx2 

the  arc  of  a  circle. 

I.  Beam  fixed  at  one  end  only,  and  loaded  with  W  at  the  other. 
Fig.  23. 

If  /o  is  the  moment  of  inertia  at  the  largest  section,  which  is 
in  this  case  at  the  wall,  and  /  is  the  variable  moment  of  inertia, 


____    __  _ 

dx*    El        EF         "    ~EI0' 

dv         Wl 

-r-=-^^#  +  C;          C=o. 

doc        EI0 

Wl 


2EIQ' 

a  deflection  50  per  cent,  in  excess  of  that  of  the  corresponding 
uniform  beam,  while  the  maximum  slope  is  twice  as  great. 


98 


STRUCTURAL  MECHANICS. 


Examples. — If  a  triangular  sheet  of  metal,  like  the  dotted  triangle 
in  Fig.  49,  is  cut  into  strips,  as  represented  by  the  dotted  lines,  and 


these  strips  are  superimposed  as  shown  above,  the  strips,  if  fixed 
at  the  ends,  and  subjected  to  W  as  shown,  will  tend  to  bend  in  arcs 
of  circles,  and  will  remain  approximately  in  contact.  If  /=io  in., 
b=4  in.,  h  =  \  in.,  W=4oo  Ib.  and  £=28,000,000,  the  deflection  will  be 

4oo-io3-43-i2 

—  -  -  =  1.37  m. 
2-28,000,000-4 

An  elliptical  steel  spring  2  ft.  long,  of  4  layers  as    shown,  each 
2  in.  broad  and  J  in.  thick,  under  a  load  ot  100  Ib.  at  its  middle,  will, 

TOO-  i23-83-  12 
if  £=29,000,000,  deflect  -  —  3=2-3  m-     The  maximum  unit 

2  -2Q,  000,000  -8 

fibre  stress  will  be  5°  12'    '     -28,800  Ib.     Note  that  one-half  of  the 

o 

weight  is  found  at  each  hinge,  and  that  the  deflection  of  one  arm  is 
doubled  by  the  use  of  two  springs  as  shown. 

II.  Beam  fixed  at  one  end  only  and  uniformly  loaded  with 
w  per  unit.     Fig.  25. 

I      b     (l-x)2         (l-x}2    I2 
I0~b0~     I2     '          ~T~  ~/o 


dx2 
dv 
dx 


2EI 

wl2 


•x  +  C; 


wl2 


v=  — 


2EIQ' 

C=o. 
C'=o. 


a  deflection  twice  that   of  the  corresponding  uniform  beam. 


FLEXURE  AND  DEFLECTION   OF  SIMPLE  BEAMS.  99 

In  these  two  cases  there  are  no  constants  of  integration,  since 

dv 

•T-  and  v=o  when  x=o. 
ax 

III.  Beam  supported  at  both  ends  and  carrying  W  at  middle. 
Fig.  27. 

I:lQ  =  b:b0=x:%l. 

d2v     Wx      Wl 


dv      Wl 

•j-=— 

dx 


Slope  =o    when    x  =  ^l;        .'.  C=  —\l. 
Wl 


w 
*w.-    32£/o, 

a  deflection  50  per  cent,  greater  than  for  a  corresponding  uniform 
beam. 

IV.  Beam   supported    at    both    ends,    and   uniformly  loaded 
with  w  per  unit  of  length.     Fig.  29. 


d2v      w  wl2 

T-9=~^T^OC~ X)  =  OT?T    > 

dx2     2Er  8£/0 

dv      wl2  _ 

" 

C'=o. 


a  deflection    20    per    cent,  greater    than    for    a   corresponding 
uniform  beam. 

98.  Uniform    Strength    and    Uniform    Breadth.  —  In    these 

h* 
cases,  as  b  is  constant,  I:Io  =  h3:ho3  or  7=/o  y-;. 


100  STRUCTURAL  MECHANICS. 

V.  Beam  fixed  at  one  end  only,  and  loaded  with  W  at  the 
other.     Fig.  24. 

£L=t£          rhz      (/~*)j 

~        '  ~  °  * 


__/      w  /      v-i 

2"     £/(/  ~£/0( 


2wii  r' 

—pj-    [(/-*)* - 

•"*  0  «^o 


or  twice  the  deflection  of  a  corresponding  uniform  beam. 

VI.  Beam  fixed  at  one  end  only  and  uniformly  loaded  with 
w  per  unit  of  length.     Fig.  26. 

h     l-x  (l-x)* 


W3 

r 


(/  log  o-/-/  log  o-/  log  /+  /  log  /)  =  - 


or  four  times  the  deflection  of  a  corresponding  uniform  beam. 

*  Log  (/—  x)=u;  dx=dv;   x=v;  du=—-——. 

//*    X  X  I 

\og(l—x)dx=x\og(l—x)+  I-  -  dx.     By  division,;  -  =  —  1+-  -  ; 
J   /  —  X  I  —  X  I  —  X 


FLEXURE  AND   DEFLECTION  OF  SIMPLE  BEAMS.         101 


VII.  Beam  supported  at  both  ends  and  carrying  W  at  middle. 
Fig.  28. 


_  T-T 

Ao2~K          :y°- 

Wfl 


dv        Wl* 


dx    4\/2jE/o 


=/       =  o  and  C  =  - 


£ 

2 

_Jo 


/2    /»          /t   \ 

\3    2^2        2A/2/ 


or  twice  the  deflection  of  a  corresponding  uniform  beam. 

VIII.  Beam  supported  at  both  ends,  and  '  uniformly  loaded 
with  w  per  unit  of  length.     Fig.  30. 


dv 
dx 

Vmax. 


^T?T    i       \  versin      ,  — 
i6EInJ0     \  I 


—  =o,  when  x  =  %l]        .'.  C=  —versin"1  i=  — 

r*  i        _2x 

JQ  ' 

r 

i  (# 

_n  _5./i  _  0.2854^/4 

oLa     4J 


wl3 
i6JE/ 


i6JB/ 


102  STRUCTURAL  MECHANICS. 

or  37  per  cent,  greater  deflection  than  for  a  corresponding  uniform 
beam. 

Other  beams  might  be  analyzed  where  both  b  and  h  varied 
at  the  same  time.  The  method  of  analysis  would  agree  with 
the  above;  but  the  cases  are  not  of  sufficient  practical  value  to 
warrant  their  discussion  here. 

99.  Sandwich  Beam. — If  a  beam  is  made  up  from  two 
materials  placed  side  by  side,  as  when  a  plate  of  steel  is  bolted 
securely  between  two  sticks  of  timber,  the  distribution  of  the  load 
between  the  several  pieces  can  be  found  from  the  consideration 
that  they  are  compelled  to  deflect  equally.  If  the  pieces  are 
of  the  same  depth,  the  extreme  fibres  of  the  two  materials  must 
undergo  the  same  change  of  length;  or  by  §  10,  if  the  subscripts 
w  and  s  stand  for  wood  and  steel  respectively, 

«i    _  j  _  fw       is  iw w 

^W  ~     S==~E^      :=  ~I?     '  =    27    * 

J^w       &s  la        J^a 

The  resisting  moment  of  such  a  beam  is 
M  =  £/„,  bw  h?  +  J/A/*2  =  II  w  (bw 

Such  beams  are  easily  figured  by  substituting  for  the  steel  an 
equivalent  breadth  of  wood  and  proceeding  as  if  the  beam  were 
entirely  of  wood  or  vice  versa. 

If  E  for  timber  is  1,400,000  and  for  steel  28,000,000,  the  ratio 
of  the  stresses  will  be  ?V;  and  if  jw  =800  lb.,  /«  =  16,000  Ib.  on  the 
square  inch. 

Example. — Two  4Xio-in.  sticks  of  timber  with  a  loXj-in.  steel 
plate  firmly  bolted  between  them  will  have  a  value  of 

„,     (800-8+  16,000-Dio2  . 

M=±—  -^ =  173^333  m.-lb., 


the  plate  supplying  T6^  of  the  amount.     The  combination  for  a  span 

A.M 

of  10  ft.  would  safely  carry  -  —=5,778  lb.  load  at  center,  or  11,555 
lb.  distributed  load,  in  place  of  3,555  or  7,110  lb.  for  the  timber  alone. 


FLEXURE   AND   DEFLECTION  OF  SIMPLE  BEAMS.         103 

100.  Resilience  of  a  Beam. — If  a  beam  carries  a  single  weight 
W,  and  the  deflection  under  that  weight  is  Vi,  the  external  work 
done  by  that  static  load  on  the  beam  is  %Wvi.  If  this  value  of 
vi  is  that  which  causes  the  maximum  safe  unit  stress  /,  the  quantity 
%Wvi  is  known  as  the  resilience  of  the  beam,  or  the  energy  of  the 
greatest  shock  which  the  beam  can  bear  without  injury,  being 
the  product  of  a  weight  into  the  height  from  which  it  must  fall 
to  produce  the  shock  in  question.  For  a  beam  supported  at 
both  ends,  loaded  in  the  middle,  and  of  rectangular  section, 

//2  2fbh? 

and        W=-^T- 

I     2     fbh*      IP  I      P 


2  3      / 

The  allowable  shock,  or  the  resilience,  is  therefore  propor- 
tioned to  }2+E,  which  is  known  as  the  modulus  of  resilience  of 
the  material,  and  to  the  volume  of  the  beam.  These  relation- 
ships hold  for  other  sections,  and  beams  loaded  and  supported 
differently.  The  above  formula  should  not  be  rigorously  applied 
to  a  drop  test,  unless  /  is  below  the  yield-point. 

Example. — A  2X2  in. -bar  of  steel,  5  ft.  between  supports,  if  /= 
16,000  lb.,  ought  not  to  be  subjected,  from  a  central  weight,  to  more 

i   i6,ooo2-22-6o  . 

than  — -  —  —  =  1 18  in.-lb.  of  energy. 

1 8     29,000,000 

If  the  load  is  distributed  over  a  similar  beam,  the  deflection 
at  each  point  will  be  v,  and  the  total  work  done  will  be 
If  w  is  uniform,  and  the  beam  is  supported  at  its  ends, 

%wf0vdx=^if0  (T  -^~l^}dx= 

Wl2      //  C      f/2 

As  -3-  =  —         and 

o      y\ 

the  above  expression  becomes 

w2l5       8 
Resilience  = ^7  =  — (wl)v\. 


104  STRUCTURAL  MECHANICS. 

Example.  —  A  weighted  wheel  of  1,000  Ib.  drops  J  in.  by  reason  of 
a  pebble  in  its  path  at  the  middle  of  a  beam  3X12  in.,  15  ft.  span. 
If  E=  1,400,000  to  find  /: 


External 


Resulting  deflection=o.69  in.  Static  unit  stress  would  be  625  lb. 
and  v=o.2  in.  In  an  actual  bridge  the  shock  is  distributed  more  01 
less  in  the  floor  and  adjacent  beams. 

101.  Internal  Work.  —  The  internal  work  done  in  a  beam  may 
be  divided  into  two  parts,  that  of  extending  and  compressing  the 
fibres,  and  that  of  distorting  them,  the  first  being  due  to  the 
bending  stresses,  and  the  second  to  the  shearing  stresses.  The 
work  of  bending  will  be  found  first. 

Let  the  cross-section  be  constant.  The  unit  stress  at  any 
point  of  a  cross-section  =  p;  the  force  on  a  layer  zdy^pzdy.  The 
elongation  or  shortening  of  a  fibre  unity  of  section  and  doc  long 
by  the  unit  stress  p  =  pdx+E.  The  work  done  in  stretching  or 

shortening  the  volume  zdydx  =  ---j=r  -zdydx.     But  p=  —  y=~ry» 

The  work  done  on  so  much  of  the  beam  as  is  included  between 
two  cross-sections  doc  apart  will  be 


fzdy-^dx. 


Substitute  the  value  of  M  for  a  particular  case  in  terms  of 
x  and  integrate  for  the  whole  length  of  the  beam.  Thus  for  a 
beam  supported  at  ends  and  loaded  with  W  at  the  middle,  M  =  %Wx 
at  any  point  distant  x  from  one  end  for  values  of  x  between  o 
and  i/.  Then 

IT72/3 


FLEXURE  AND  DEFLECTION  OF  SIMPLE  BEAMS.         105 

The  internal  work  due  to  shear  is  not  readily  found  unless 
a  simple  form  of  cross-section  is  chosen.  If  the  section  is  rect- 
angular the  shear  at  any  point  on  the  cross-section  is,  by  §  72, 


The  distortion  in  a  length  doc  is  qdoc+C  and  the  work  done  upon 

g    qdoc 
a  volume  zdydx  is  —  •  -j=r  •  zdy.     The  work  done  upon  the  volumes 

2        O 

between  two  sections  dx  apart  is 
dx    r+^h 


For  a  beam  carrying  a  single  load  at  the  centre,  as  above,  F 
a  constant,  and  the  work  done  by  the  shearing  forces  is 


20  bhC 

102.  Deflection  Due  to  Shear.  —  In  the  cases  of  flexure  so  far 
treated  the  bending  moment  only  has  been  taken  into  account 
in  finding  the  deflection.  The  shearing  stresses,  however, 
cause  an  additional  deflection  which  is  generally  too  small  to  be 
of  practical  account.  The  deflection  of  a  rectangular  beam 
carrying  a  single  load  at  the  centre  can  be  found  from  the  results 
of  the  preceding  section  since  the  internal  work  must  be  equal 
to  the  external. 


WP 


The  first  of  these  terms  is  the  deflection  when  the  shear  is  not 
taken  into  account,  and  it  has  the  same  value  as  was  obtained 

in  §  94.     The  ratio  of  the  second  term  to  the  first  is    r2  ,  which 

5C/ 

shows   that  in  rectangular  beams   of  ordinary  proportions   the 
shearing  deflection  is  but  a  small  proportion  of  the  whole. 


106  STRUCTURAL  MECHANICS. 

Examples. — i.  What  is  the  deflection  at  the  middle  of  a  2Xi2-in. 
pine  joist  of  12  ft.  =  144  in.  span,  supported  at  ends  and  uniformly 
loaded  with  3,200  Ib.  ?  £=1,600,000.  0.27  in. 

2.  What  is  the  deflection  if  the  load  is  at  the  middle?       0.432  in. 

3.  Find  the  stiff est  rectangular  cross-section,  bh,  to  be  obtained 
from  a  round  log  of  diameter  d.  b=$d. 

4.  A  4X6-in.  joist   laid  flatwise  on  supports  10  ft.  apart  is  loaded 
with  1,000  Ib.  at  the  middle.     The  deflection  is  found  to  be  0.7  in. 
What  is  E?  1,607,000. 

5.  What  is  the  maximum  safe  deflection  of  a  i2-in.  floor-joist,  14  ft. 
span,  if  /=i,2oo  Ib.  and  £=1,600,000? 

0.37  in.  for  uniform  load;  0.29  in.  for  load  at  middle. 

6.  What  is  the   diameter  of  the  smallest  circle   into  which  J-in. 
steel  wire  can  be  coiled  without  exceeding  a  fibre  stress  of  20,000  Ib. 
per  sq.  in.  ? 


CHAPTER  VII. 
RESTRAINED  AND  CONTINUOUS  BEAMS. 

103.  Restrained  Beams.  —  When  a  beam  is  kept  from  rotating 
at  one  or  both  points  of  support,  by  being  built  into  a  wall,  or 
by  the  application  of  a  moment  of  such  a  magnitude  that  the 
tangent  to  the    curve    of   the    neutral    plane    at    the   point    of 
support  is  forced  to  remain  in  its  original  direction  (commonly 
horizontal)  at  such   point,  the  beam  is  termed  fixed  at  one  or 
both  supports.     The  magnitude  of  the  moment  at  the  point  of 
support  depends  upon  the  span,  the  load  and  its  position.     It 
is  the  existence  of  this  at  present  unknown  moment  which  calls 
for  the  application  of  deflection  equations  to  the  solution  of  such 
problems  as  those  which  follow,  there  being  too  many  unknown 
quantities  to  permit  the  treatment  of  Chapter  III. 

In  applying  the  results  obtained  in  the  following  cases  to  actual 
problems,  one  should  feel  sure  that  the  beam  is  definitely  fixed 
in  direction  at  the  given  point.  Otherwise  the  values  of  M,  F, 
and  v  will  be  only  approximately  true. 

104.  Beam  of  Span  /,  Carrying  a  Single  Weight  W  in  the 
middle  and  supported  and  fixed  at  both  ends.     Origin  at  left 
support.     Fig.  50. 

The  reactions  and  end  moments  are  now  unknown.  The 
beam  may  be  considered  either  as  built  in  at  its  ends  (as  at  the 
right  in  above  figure),  or  as  having  an  unknown  couple  or  moment 
Qb  applied  at  each  point  of  support  (as  at  the  left),  of  a  magnitude 
just  sufficient  to  keep  the  tangent  there  horizontal. 

The  reaction  at  either  end  will  then  be  %W  +  Q,  while  the 
shear  between  the  points  of  support  will  still  be±}J/F.  For 
values  of  x  <  J/, 


107 


io8 


STRUCTURAL  MECHANICS. 


If  this  value  is  compared  with  that  of  Mx  in  §  94,  it  is  seen 
that  a  constant  subtractive  or  negative  moment  is  now  felt  over 
the  whole  span  in  combination  with  the  usual 


Slope  is  zero  when  #  =  o;     /.  C  =  o.     Also   slope   is   zero  when 
#-#. 

o  =  ^Wl2  -  IQbl.         -Qb=-  \Wl, 

the  negative  bending  moment  at  either  end.     That  it  is  negative 
appears  by  making  oc=o  in  Mx  above. 

If  this  value  of  Qb  is  substituted  in  the  first  equation,  giving 
Mx  =  %W(x—  i/),  the  point  of  contraflexure  is  located  at  #=J/, 
and  the  bending  moment  at  middle,  where  x=%l,  is  -M™^ 
=  +JF7,  or  one-half  the  amount  in  §  94.  Substituting  the  value 
of  Qb  in  the  equation  for  slope, 


dv 


W 

4E/( 


W 


As  v=o  when  x  = 


=  o.     When#=|/, 


RESTRAINED  AND  CONTINUOUS  BEAMS.  109 

W    /I3       13\  Wl3 


Vy 


IQ2E/' 


The  beam  is  therefore  jour  times  as  stiff  as  when  only  sup- 
ported at  ends. 

//      Wl,-       8/7  //2 

As   —  =  -3-'  W  =  —  -r      and     Vi        ' 
i       8  i/ 


so  that  only  one-half  the  deflection  is  allowable  that  is  permitted 
in  §  94,  but  the  beam  may  safely  carry  twice  the  load. 

It  is  useful  to  notice  that  this  beam  has  a  bending  moment 
at  the  middle  equal  to  that  which  would  exist  there  if  the  beam 
were  cut  at  the  points  of  contraflexure  and  simply  supported 
at  those  points,  and  that  the  two  end  segments,  of  length  J/, 
act  like  two  cantilevers  each  carrying  JTF,  the  shear  at  the  point 
of  contraflexure. 

If  the  weight  were  not  at  the  middle  the  moments  at  the 
two  ends  would  differ,  equations  would  be  needed  for  each  of 
the  two  segments,  and  the  solution,  while  possible,  would  be 
much  more  complicated. 

Example.  —  A  wooden  beam,  6  in.  square  and  7}  ft.  span,  is  built 
into  the  wall  at  both  ends.  A  central  weight  of  3,000  Ib.  will  give  a  max. 

fibre  stress  of  ^—  -•  -^  —  =  937^  Ib.  per  sq.  in.  at  the  middle  and  both 

2 

ends.   The  deflection  will  be  —  —  -  -  -  -r=  0.07  in.  if  E=  i  ,  qoo.ooo. 

192  -i,  500,000  -64 

The  allowable  deflection  for  /=  1,200  is  —  -  -  =  0.00  in.,  and 

24-1,500,000-3 

.max.  allowable  W=  3,  006-  4=3,  860  Ib. 

105.  Beam  of  Span  /,  Uniform  Load  of  w  per  unit  over  the 
whole  span,  fixed  at  both  ends.  Origin  at  left  support.  Fig.  51. 

As  in  the  previous  case,  the  reaction  at  either  end  may  be 
represented  by  %wl+Q.  The  shear  at  x  is  %wl—wx,  which  ex- 
pression changes  sign  at  the  middle  and  at  either  point  of  support; 
hence  at  those  places  will  be  found  MmaXt 


HO  STRUCTURAL  MECHANICS. 

Compare  with  §  93. 


dv 

•j~=o  when  #  =  o;    . 

ivl3    wl3 
., 


dv 
C=o.        -r-  =  o  when 


wl2 
-, 


•Fig.  51 


the  negative  moment  at  each  point  of  support.     If  x 

M  at  middle  = 
Substitute  the  value  of  Qb  and  get 


wl2 

-  1  -TV)  =  — 
24 


#*       &?      JW_ 

12         6  12 


RESTRAINED   AND   CONTINUOUS  BEAMS. 
Since  ^  =  o    when  x = o,     C'  =  o.     When  x  =  J/, 


Ill 


which  is  one-fifth  the  value  of  §  93. 

The  points  of  contraflexure  occur  where  Mx  —  omt 


The  second  term  is  the  distance  from  the  middle,  each  way, 
to  the  points  of  contraflexure.  If  M  is  calculated  for  the  middle 
point  of  a  span  /-^-\/3,  it  will  prove  to  be  wl2-i-24,  as  above. 
Since 


(wT)lI 


12/7 


12 


and 


or  0.3  as  much  as  in  §  93.     The  beam  may  safely  carry,  however, 
50  per  cent,  more  load. 

Example. — An  8-in.  I  beam  of  12  ft.  span,  carrying  1,000  Ib.  per  foot, 
if  firmly  fixed  at  both  ends    and  not  to  have  a  larger  unit  stress  than 

,,,,,,  .    T     i,  ooo  -12  -12  -12  -4 

12,000  Ib.,  should    have  a  value  of  /= —  —  =  48.     An 

12-12,000 

8-in.  steel  beam,  18  Ib.  to  the  foot,  7=  57.8,  will  satisfy  the  requirement, 
the  load  then  being  1,018  Ib.  per  foot.     The  deflection  will  be  0.06  in. 


Fig.  53 


106.  Beam  of  Span  /,  fixed  or  horizontal  at  P2,  supported 
at  PI  and  carrying  a  uniform  load  of  wper  unit.  Fig.  52.  Origin 
at  PI  and  reaction  unknown. 


112  STRUCTURAL  MECHANICS. 

It  will  be  seen  from  the  sketch  that  a  beam  of  length  2/,  rest- 
ing upon  three  equidistant  supports  in  the  same  straight  line,  will 
come  under  this  case. 


- 
dx 


i  /Fix*    woe4     wPx     PiPx        \ 
v=—  -  —  2  --  --  +  -7  --  --  +  C'I. 
EI\    6         24        6  2  / 

T;=O    when  #  =  o;     .'.^  =  0.     If  oc=lt     v  =  o,     and 


Substitute  this  value  of  PI  in  the  above  equations. 


As   a   rninimum  value   of  v,  or  v  =  o,  occurs  for  x=l,  divide 
by  x—  I  and  obtain 

Sx2-lx-P  =  o         or        x 


Then  *W.  =  -0.0054 


^-. 

To  find  points  of  Mmax,  put  Fx=o,  or  x=%l. 
Also,  by  inspection,  M^a^  when  ^=/. 
For  *  =  §/,  Mwo,.  =(A 


RESTRAINED   AND   CONTINUOUS  BEAMS.  113 


For  *=/,  Mmax.=($ 

For  the  point  of  contraflexure  \lx  —  J#2=o,  or  #=f/,  as 
was  to  be  expected  from  the  position  of  the  point  of  maximum 
positive  M. 

Note  again  that  the  point  of  maximum  bending  moment 
is  not  the  point  of  maximum  deflection. 

It  will  be  seen  that  a  continuous  beam  of  two  equal  spans  /, 
uniformly  loaded  with  w  per  unit,  has  end  reactions  of  f  wl  and 
a  central  reaction  of  2Xfw/  =  f'ie>/;  that  points  of  contraflexure 
divide  each  span  at  \l  from  the  middle  pier,  and  that  the  bending 
moment  at  the  middle  of  the  remaining  segment  of  f  /  is,  as  above, 
f  -%-^wl2  =  j%-%wl2.  It  will  also  be  seen  that,  since  the  bending 
moment  at  P^  is—  J-w/2,  a  uniform  beam,  continuous  over  two 
equal  spans,  each  /,  is  no  stronger  than  the  same  beam  of  span 
/  with  the  same  uniform  load.  It  is,  however,  about  two  and  a 
half  times  as  stiff. 

Example.  —  A  girder  spanning  two  equal  openings  of  15  ft.  and 
carrying  a  i6-in.  brick  wall  10  ft.  high  of  no  Ib.  per  cubic  ft.  will 

throw  a  load  of  —  -  =27,500   Ib.  on  the  middle  post  and 

f  4-  no-  10-  iq-  is 

must  resist  a  bending  moment  of  -  —  -  —  ^=41,250  ft.-lb. 

3  *  " 

107.  Two-span    Beam,    with     Middle    Support     Lowered.  — 
A  uniform  beam,  uniformly  loaded  and   supported  at  its  ends, 
will  have  a  certain  deflection  at  the  middle  which  can  be  calculated. 
If  the  middle  point  is  then  lifted  by  a  jack  until  returned  to  the 
straight  line  through  the  two  end  supports,  the  pressure  on  the 
jack,  by  §  106,  will  be  five-eighths  of  the  load  on  the  beam.     Since 
deflection  is  proportional  to  the  weight,  other  things  being  equal, 
if  the  jack  is  then  lowered  one-fifth  of  the  first  deflection  referred 
to,  the  pressure  on  the  jack  will  be  reduced  one-fifth,  or  to  one-half 
of  the  load  on  the  beam.     Hence,  if  a  uniformly  loaded  beam  of 
two  equal,  continuous  spans  has  its  middle  support  lower  than 
those  at  its  ends  by  one-fifth  of  the  above  deflection,  the  middle 
reaction  will  be  one-half  the  whole  weight,  the  bending  moment 
will  be  zero  at  the  middle,  and  the  beam  may  be  cut  at  that  point 
without  disturbance  of  the  forces. 

108.  Beam  of  Span  /,  fixed  at  left  and  supported  at  right  end, 


STRUCTURAL  MECHANICS. 


and  carrying  a  single  weight  W  at  a  distance  a  from  the  fixed  end, 
Fig.  53.     Origin  at  fixed  end. 


.Fig.  53 


The  reaction  at  the  supported  end,  being  at  present  unknown, 
will  be  denoted  by  P2,  and  moments  will  be  taken  on  the  right  of 
any  section  x.  From  lack  of  symmetry,  separate  expressions 
must  be  written  for  segments  on  either  side  of  W. 


BETWEEN    W   AND   FIXED    END. 
Mx  =  P2(l -  *)  -  W(a  -  X) 


El 


+  Cx+C' 


dv 

— =o  when  x=o;     .'.  C=o. 

dx 

v=o  when  x=om,     .*.  C'=o. 

.-.  C"  =  -W(a2- 


If 

or 


BETWEEN  W  AND  SUPPORTED  END. 


If  x=a,  —  on  left =3-  on  right. 
'  dx  dx 

If  x=a,  v  on  left=v  on  right. 


Wa2 
-Ja)  -^i/3  — - 


RESTRAINED  AND  CONTINUOUS  BEAMS.  US 

If  this  value  of  P^  is  substituted  in  the  above  equations  the  desired 
expressions  are  obtained.     Thus 


Mtnax.,  by  inspection,  when  oc=o,    or    x=a. 

Wa2  W 

Mmax.  =  -^(^l-^-Wa  =  —-pa(l-a)(2l-a)  at  fixed  end, 

Wa2 
and  =—  w-(l—  aX^~  a)  at  tne  weight. 

The  point  of  contraflexure  occurs  between  W  and  the  fixed 
end,  where  M  =  o,  or 


The  maximum  deflection  will  be  found  where  -j-  =  o,  on  the 

d# 

right  or  left,  according  to  the  value  of  a. 

The  above  beam  may  be  regarded  in  the  light  of  two  equal 
continuous  spans  with  W  on  each,  distant  a  each  side  of  the 
middle  point  of  support. 

In  solving  the  more  intricate  problems  in  the  flexure  of  beams, 
as  well  as  those  just  treated,  each  equation  of  condition  can  be 
used  but  once  in  the  same  problem,  and  as  many  unknown 
quantities  can  be  determined  as  there  are  independent  equations 
of  condition.  The  reactions  and  moments  at  the  points  of  support 
are  usually  unknown,  and  must  be  found  by  the  aid  of  such 
flexure  equations  as  have  just  been  used. 

Example.  —  A  bridge  stringer  which  is  continuous  over  two 
successive  openings  of  12  ft.  each  and  carries  a  weight  from  the 
wheels  of  a  wagon  of  3,000  Ib.  at  each  side  of  and  3  ft.  from  the 
middle  support  will  be  horizontal  over  that  support.  Then  —  MmaXt 

=  -S-3-9"!  =  -  5,9°6.25    ft,lb.      +  Mmax=^3.32.33.9 
=  2,320.3  ft.-lb.    P2=3'0003-32'33=258  lb-     Reaction  at  middle  sup- 


port from  both  spans  =  2  (3  ,000—  2  58)  =5,  484  Ib. 


n6 


STRUCTURAL   MECHANICS. 


109.  Clapeyron's  Formula,  or  the  Three-moment  Theorem 
for  Continuous  Loading. — To  find  the  reactions,  shears,  and 
bending  moments  for  a  horizontal  uniform  continuous  beam 
loaded  with  wi}  w2>  w3j  etc.,  loads  per  running  unit  over  the 
successive  spans  h,  ht  Is,  etc.  Fig.  54.  P0,  PI,  P2,  etc.,  denote 


Fig,  54 


the  unknown  reactions;  M0,  MI,  M2,  etc.,  the  unknown  bend- 
ing moments  at  points  of  support,  o,  i,  2,  etc.;  FQ,  FI,  F2,  etc., 
denote  the  shears  immediately  to  the  right  of  o,  i,  2,  etc.;  while 
Fit  FJt  etc.,  denote  the  shears  immediately  to  the  left  of  the 
points  of  support  i,  2,  etc. 

The  origin  of  coordinates  is  first  taken  at  O  and  the  sup- 
ports are  on  a  level.  +M  makes  the  beam  concave  on  the  upper 
side.  As  positive  shear  acts  upward  at  the  left  of  any  cross- 
section,  w  is  negative. 

Consider  the  condition  of  equilibrium  of  the  first  span  o-i, 
or  /i,  loaded  throughout,  with  w\  per  unit  of  length. 

Take  moments  on  the  left  side  of  and  about  a  section  S, 
distant  x  from  the  origin  O.  The  bending  moment  at  S  is 


RESTRAINED   AND   CONTINUOUS  BEAMS.  "7 

.  .  .  .  (i) 


Let  i0j  ii9  i2*..in  =  tangent  of  inclination  of  the  neutral 
axis  at  o,  i,  2,  .  .  .  N.  Integrate  (i)  and  transpose  the  con- 
stant of  integration,  z'o,  to  the  left-hand  member  and  thus  obtain 
an  expression  for  the  difference  in  slope  or  inclination  of  the 
two  tangents  to  the  bent  beam  at  O  and  S. 


(2) 


dv 
When  x—li,  T~=^i>  and  hence 

Elfa-id^MJi  +  lFotf-frvih*.      ...    (3) 

Integrate  (2)  and    determine  constant  as  zero,  because  v=o 
when  x=o. 


Make  #=/i,  then  V=VI=Q} 
and  -EH0li  =  JM0/ 

or  -E/*0  =  iMo/i  +  t^o/i2-Awi/i8  .....    (5) 

Eliminate  i0  by  subtracting  (5)  from  (3). 

.....     (6) 


If  the  origin  is  taken  at  i  instead  of  o  an  equation  like  (5) 
is  obtained  for  the  second  span  /2>  or 


.....     (7) 

Add  (6)  and  (7),  obtaining 

O  =  P/  o/i  +  Pf  !/2  +  Wl2  +  i^l/22  -  JWl/l3  -  AW2/28-          (8) 


Il8  STRUCTURAL  MECHANICS. 

The  unknown  slopes  have  thus  been  eliminated.  The  next 
step  is  to  remove  either  M  or  F.  Equation  (i)  must  equal  MI, 
for  x = l\ .  Therefore 


Mi  =  Mo  +  ^Vi  —  ^vvili2,       or      FQ 

In  the  same  way,  for  second  span,  -Fi  =  — — -, — 
Substitute  the  values  in  (8)  and  obtain 

Mi/2     MI— M0       wJi3    M2—  MI 

-\ 1 /i  H — 7 1- ' 


223  66 

W2/23       Wi/i3        W2/23 

+  '      7)  • 

12          8  24 

.'.  MO/I  +  2Mi  (/i  +  /2)  +  M2/2  =  -  iOi/i3  +  Wa/28),  •     •    (9) 

which  is  Clapeyron's  formula  for  pier  moments  for  a  continuous 
beam,  with  continuous  load,  uniform  per  span.  Notice  the 
symmetry  of  the  expression.  The  negative  sign  to  the  second 
member  indicates  that  the  bending  moments  at  points  of  sup- 
port are  usually  negative. 

Example.  —  Three  spans,  30  ft.,  60  ft.,  and  30  ft.  in  succession. 
Load  on  first  and  last  500  Ib.  per  ft.,  on  middle  span  300  Ib.  per  ft. 
No  moment  at  either  outer  end.  Then  MQ=O.  M\  =  M%  by  sym- 
metry. 

2M  i  •  90+  M2  •  60=  —  t(5oo  •  3O3+  300  •  6o3). 
Mi=  -81,562$  ft.-lb.        F0=  -2,7181+7,500=  +4,78iJ  Ib. 
FI'=  +4,7811—30-500=  —  10,218}  Ib.        FI  =  300  -30=  9,000  Ib. 
/.  P0=^4=4,78iilb.;        Pi  =  P2=io,2i8i+9,ooo=i9,2i8ilb.  ^ 
The  bending  moment  and  shear  at  any  point  can  now  be  readily 
determined. 

If  the  two  adjacent  spans  are  equal  and  have  the  same  load, 
M0+4Mi+M2=  -%wl2  ......    (10) 


If  there  are  n  spans,  n  —  i  equations  can  be  written  between 
n  +  i  quantities  MO,  MI  .  .  .  Mn.  But  if  the  beam  is  simply 
placed  on  the  points  of  support,  the  extremities  being  unre- 
strained, MO=O  and  Mn=o,  and  there  remain  n  —  i  equations 


RESTRAINED   AND   CONTINUOUS  BEAMS.  lip 

to  determine  MI  .  .  .  Mn-i-     If  the  beam  is  fixed  at  the  ends, 
the  equations  io  =  o  and  in  —  o  will  complete  the  required  number, 
no.  Shears  and  Reactions.  —  As  the  shear  is  the  first  derivative 
of  the  bending  moment,  §  56,  from  (i)  is  obtained 

dM 

—=F=FQ 


as  was  to  be  expected,  +F  acting  upwards  on  the  left  of  the 
section.  A  similar  equation  can  be  written  for  each  span. 

The  reaction  at  any  point  of  support  will  be  equal  to  the 
shear  on  its  right  plus  that  on  its  left  with  the  sign  reversed. 
As  the  shear  on  its  left  is  usually  negative,  the  arithmetical  sum 
of  Fn  and  Fnf  commonly  gives  the  reaction. 

A  simple  example  may  make  the  application  plainer.  Given 
two  equal  spans  on  three  supports, 

Wi=w2  =  lw.     MQ=O,  M2=o.     (10)  gives  Mi= 
=  }  wl;        Fl'  =  l<wl-wl=  - 


i  wl3 

(5)  gives  i0=  -— 


4&EP 

fr\  -L  -J 1 W73  =  / 

El 

and  the  analogous  equation  for  the  second  span  is 


(6)  gives  ii  =  -^7(0  +  ^-  -  J) wl3  =  o, 


which  differs  from  io  only  in  direction  of  slope. 

dv        wl3  woe3 


wl3  woe* 

(4)  gives  EIv  =    ---goc  +?\wlx3  -  -  — 


120  STRUCTURAL  MECHANICS. 

These  equations  determine  the  slope  and  deflection  at  each 
point.     Putting  ~J~=°>  there  results  l^—glx2-\-Sx3=ot   containing 

the  root  x=l,  already  known.  Therefore  divide  by  l—oc  and 
obtain  l2+lx  —  8#2=o,  which  is  satisfied  for  ^=0.4215^,  the  point 
of  maximum  deflection.  The  substitution  of  this  value  in  the 
equation  for  v  will  yield  vniaXm 

From  (i)  M  =  %wlx  —  %wx2. 

If  M=o,  |/  —  J#=o,  or  x  =  f/,  the  point  of  contraflexure. 
Differentiate  M  and  get  F  =  %wl—  wx. 
If  F=o,  x  =  %l,  the  point  of  +  Mmax. 


Example.  —  If   a   uniformly   loaded    continuous    beam   covers   five 
equal  spans, 


Then  M,  =  -  &wl2=  M^M^-  ^wl2=  M3. 


The  sum  of  the  reactions  must  equal  $wl. 

in.  Coefficients  for  Moments  and  Shears.  —  It  has  been 
found  that  the  numerical  coefficients  for  moments  and  shears 
at  the  points  of  support,  when  all  spans  are  equal  and  the  load 
is  uniform  throughout,  may  be  tabulated  easily  for  reference  and 
use.  Thus  the  values  of  M  and  F  just  obtained  for  the  five  equal 
spans  can  be  selected  from  the  lines  marked  V.  The  reactions 
are  given  by  the  arithmetical  addition  of  the  shears.  The  sum 
of  the  reactions  must  equal  the  total  load.  The  shears  at  the 
two  ends  of  any  span  differ  by  the  whole  load  on  the  span,  the 
shear  at  the  right  end  being  negative.  The  dashes  represent 
the  spans. 


RESTRAINED  AND  CONTINUOUS  BEAMS.  121 

SHEAR  AND   REACTION   COEFFICIENTS. 

I.  i-Jtrf; 
II.  f-f  f—  i*rf; 
III.  A—  A  A—  A  A—  A; 

IV-  if-  il  if-if  M-M  H-li; 
V.  if-H  ft-it  if—  it  if-  ft  tt-tt; 

etc.,  etc. 

PIER  MOMENT   COEFFICIENTS. 

II.  -i- 
III.  — 

v.  —  A—  A—  A-A—  ; 

etc.,  etc. 

The  rule  for  writing  either  table  is  as  follows:  For  an  even 
number  of  spans,  the  numbers  in  any  horizontal  line  are  obtained 
by  multiplying  the  fraction  above,  in  any  diagonal  row,  both 
numerator  and  denominator,  by  two,  and  adding  the  numerator 
and  denominator  of  the  preceding  fraction.  Thus,  in  the  first 

2X6+  5     17  2X  i  +  i      3 

table,  —  —  —    -3—  -77,   and  in   the  second   table,   —r-  —5  =-3, 

2X10-1-8     28  2X10  +  8     28 

or  -  -^=  -  .     For  an  odd  number  of  spans,  add  the  two 

2X38  +  28     104 

preceding  fractions   in   the   same   diagonal   row,   numerator   to 

numerator  and  denominator  to  denominator.    Thus,  ~  -  -—-77. 

28  +  10    38 

The  denominators  agree  in  both  tables.     A  recollection  of  two 
or  three  quantities  will  enable  one  to  write  all  the  others. 

Example.  —  Continuous  beam  of  five  equal  spans,  each  /,  carrying 
w  per  foot.  Where  and  what  is  the  max.  +  M  in  second  span  ?  Shear 
changes  sign  at  f  §/  from  left  end  of  span.  If  this  span  were  inde- 

pendent,   +  M  at   that  point   would   be   Iwl220  * 


. 

19-19     361     8 

negative   or  subtractive   moment   is    (sV+*V*if)^2-     The   difference 
between  these  values  is  +Mmax- 


122  STRUCTURAL  MECHANICS. 

A  more  general  investigation  will  produce  equations  which 
are  of  great  practical  value  in  the  solution  of  problems  concerning 
continuous  bridges,  swing-bridges,  etc.,  as  follows  : 

112.  Three-moment  Theorem  for  a  Single  Weight.  —  O  is  the 
origin,  Fig.  55;  the  supports  are  at  distances  h  below  the  axis  of 
X.  A  single  weight  Wn  is  distant  kln  from  O  on  the  span  /n, 
k  being  a  fraction,  less  than  unity,  of  the  span  in  which  W  is 
situated. 

The  moment  at  section  S  beyond  Wn  will  be,  as  in  the  former 
discussion, 


-Wn(x-kln)  .....         (l) 

If  oc=ln,  Mx  =  Mn+i,  and  from  (i) 


_  TT7  J     Z7          Mm+i  —Mm 

For  an  unloaded  span,  W  =o,  and  Fm  =  —  —,  —    —  . 

^m 

For  the  shear  on  the  left  of  a  section  at  the  right  end  of  the 
wth  span, 


/n 

For  an  unloaded  span,  W=o,  and 

77  /     Mm—  Mm-\ 

•F  m   ~  J  '• 

As  Fmr  is  the  shear  at  left  of  support  m,  and  Fm  is  the  sheai 
at  right  of  the  same  support;  the  reaction  there  will  be  the  sum 
of  Fm  and  -Fm  or 

Mm+i  -Mm      Mm^i  —Mm 
•Lm  7  7 

I'm  "m—l. 

To  get  values  of  i  and  v  for  span  ln  it  is  necessary  to  write 
separate  equations  for  the  two  segments  into  which  Wn  divides 


RESTRAINED  AND  CONTINUOUS  BEAMS.  123 

the  span.    Equation  (i)  applies  to  the  right  segment,  and  by 
omitting  the  last  term  the  equation  for  the  left  segment  is  obtained. 

M         d2v 
Equate  ^rr  to  -7—2  and  integrate  between   the   limits  of  o  and 


***! 


**.' 


Fig.  55 


&/„  on  the  left  and  between  £/n  and  #  on  the  right.     Then  on 
the  left 

/kin  rkln 

dx+Fnl      xdx,      .    .    (2) 

and  on  the  right 
dv 


\  Cx  Cx  Cx 

ikln)=Mn    I      dx  +  Fn  xdx~W  n    I      (x-kln)dx. 

U  Kin  J  kin  «7  kin, 


(3) 


The  sum  of  (2)  and  (3)  gives  the  change  of  slope  between 
the  support  n  and  the  section  S. 


(4) 


124  STRUCTURAL  MECHANICS. 

Integrate  the  last  equation  between  the  same  limits  as  before. 
EI(v  —lnx — hn)  =  %Mnoc2  +  %Fnx3  —  %Wn(x — kln)3,      •     (5) 

which  is  the  general  equation  of  the  curve  of  the  neutral  axis, 
the  term  in  W  disappearing  for  values  of  x  less  than  kln. 

If  x=ln,  v  =  hn+i.      If  the  value  of  Fn  from  (ia)  is   inserted 
in  (5),  the  slope  at  support  n  is 

~-^(2Mnln  +  Mn+lln+Wnln2(2k-3k2  +  k3)].        (6) 

The  equation  of  the  curve  is  therefore  completely  deter- 
mi  ed  when  Mn  and  Mn+i  are  known.  The  equation  of  this 
curve,  between  Wn  and  the  w  +  ith  support  is  given  by  (5),  and 
the  tangent  of  its  angle  with  the  axis  of  X  by  (4).  If  the  value 
of  Fn  from  (ia)  and  of  in  from  (6)  are  substituted  in  (4),  and 

dv 

x=ln,  -7-  will  be  the  tangent  in+i  at 
doc 


Or 


Remove  the  origin  from  O  to  N,  and  dfrive  an  expression 
for  in  by  diminishing  the  indices. 


ln-!        ^6EI1' 

Equate  with  (6)  and  transpose. 

=6El(kn~ll~kn+tn+ll~   nj  -Wn- 


which  is  the  most  general  form  of  the  three-moment  theorem 
for  a  girder  of  constant  cross-section.  In  using  the  theorem 
the  following  factors  may  be  of  service: 


RESTRAINED  AND  CONTINUOUS  BEAMS.  125 

Pier  moments  are  usually  negative  and  the  end  moments 
zero.  When  the  supports  are  on  a  level,  hi  =  h2,  etc.,  and  the 
term  containing  El  disappears. 

Any  reaction  Pn  =  Fn-Fnf\ 


It  should  be  borne  in  mind  that  all  the  preceding  deflection 
formulas  are  derived  on  the  assumption  that  the  deflection  is 
due  entirely  to  bending  moment,  the  deformation  from  shear 
being  neglected.  In  a  solid  beam  the  amount  of  deflection  due 
to  shear  is  very  small,  but  such  is  not  the  .case  with  a  truss.  In 
a  truss  the  deflection  due  to  the  deformation  of  the  web  may  in 
some  cases  amount  to  half  the  total  deflection.  For  this  reason 
the  deflection  formulas  and  the  three-moment  theorem  should 
not  be  applied  to  trusses. 

Example.  —  Three-span  continuous  girder  carrying  loads  as  shown. 
Supports  on  a  level. 

O  3</  <N~  I    2C/  M"  40'  ro  40'  2  _  3_ 

A  A  A  A 

5°'  I00'  75' 

2  •  1  50^  +  lOolf  2  =  —  2  ,000  •  2  ,  500  •  f  •  |  •  |  -  1  ,000  •  10,000  •  TV  '  TV  '  It 

-  3  ,000  •  1  0,000  •  T6-o  •  ro  •  i  $  • 

iooMi  +  2.175^2=  -  1,000-  10,000-  fV  •  iV  If  ~3>ooo  •  1  0,000  -TV  TV  t% 
$ooM  i+  looM  2=  —  14,880,000,         TooM1  +  35oJM"2==  —  13,440,000. 
M}  =  -40,670  ft.-lb.         M2=  —  26,780  ft.-lb. 


-+11=_ 
50  50 


.3 

4-  4> 


_  -26,780+40,670  ,  40,670     1,000-8     3,000-4  ,  2,000. 
•L  i  — 

100  50  10  10  5 

+  26,780     -40,670+26,780     1,000-2     3,000-6 

2==  ~    ==~r2,2IolD. 

10  10 


75  100 

26,780 
^     =-357^ 


75 

—  i3  +  4>I52  +  2>2l8~357  =  2>000+I>000+3j°00==  6,000. 


126  STRUCTURAL  MECHANICS. 

Examples.  —  i.  A  brick  wall  16  in.  thick,  12  ft.  high,  and  32  ft. 
long,  weighing  108  Ibs.  per  cubic  ft.,  is  carried  on  a  beam  supported 
by  four  columns,  one  at  each  end,  and  one  8  ft.  from  each  end.  Find 
M  at  the  two  middle  columns  and  the  reactions. 

M-—  31,104  ft.-lb.;  PI  =  3,024  lb.;  P2=  24,624  Ib. 

2.  Two  successive  openings  of  8  ft.  each  are  to  be  spanned.     Which 
will  be  stronger  for  a  uniform  load,  two  8  ft.  joists  end  to  end  or  one 
1  6  ft.  long?    Find  their  relative  stiffness. 

3.  A  beam  of  three   equal  spans  carries  a  single  weight.     What 
will  be  the  reactions  and  their  signs  at  the  third  and  fourth  points 
of  support  when  W  is  in  the  middle  of  the  first  span? 


4.  A  beam  loaded  with  50  lb.  per  foot  rests  on  two  supports  15  ft. 
apart  and  projects  5  ft.  beyond  at  one  end.     What  additional  weight 
must  be  applied  to  that  .end  to  make  the  beam  horizontal  at  the  nearer 
point  of  support?  i$6i  lb.  at  the  end,  or  312  J  lb.  distributed. 

5.  A  beam  of  two  equal  spans  on  level  supports  carries  a  single 
load,  W,  in  the  left  span.     Prove  that 


CHAPTER  VIII. 
PIECES  UNDER  TENSION. 

113.  Central  Pull. — If  the  resultant  tension  P  acts  along  the 
axis  of  the  piece,  the  stress  may  be  considered  as  uniformly  dis- 
tributed on  the  cross-section  S.     If,  then,  /  is  the  maximum  safe 
working  stress  per  square  inch  for  the  kind  of  load  which  causes 
P,  Fig.  56, 

P  =  fS,     or     S=j 

for  the  necessary  section  which  need  not  be  exceeded  throughout 
that  portion  of  the  piece  where  the  above  conditions  apply. 
Changes  due  to  connections  will  require  a  larger  section. 

If  owing  to  lack  of  uniformity  in  the  material  or  the  direct 
application  of  P  at  the  end  of  a  wide  bar  to  a  limited  portion 
only  of  the  width  the  stress  may  not  be  considered  as  uniformly 
distributed  at  a  particular  cross-section,  injurious  stress  may  be 
prevented  by  taking  the  mean  stress  /  at  a  smaller  value  and 
obtaining  a  larger  cross-section. 

If  there  is  lack  of  homogeneity,  or  two  materials  are  used 
together,  or  two  or  more  bars  work  side  by  side,  those  fibres 
which  offer  the  greatest  resistance  to  stretching  will  be  subject  to 
the  greatest  stress.  Fortunately  the  slight  yielding  and  bending 
of  connecting  parts  tend  to  restore  equality  of  action. 

A  long  tension  member  has  a  much  greater  resisting  power 
against  shock  than  a  short  one  of  equal  strength  per  square  inch. 
See  §  20. 

114.  Eccentric  Pull. — If  the  variation  of  stress  on  a  cross- 
section  is  due  to  the  fact  that  the  line  of  action  of  the  applied 

127 


128 


STRUCTURAL  MECHANICS. 


force  does  not  traverse  the  centre  of  figure  of  the  cross-section 
S,  the  force  P'  that  can  be  imposed  without  causing  a  unit  stress 
greater  than  /  at  any  point  in  the  section  is  less  than  P  of  the  pre- 
ceding formula,  and  depends  upon  the  perpendicular  distance 
yo  of  the  action  line  of  P  from  the  centre  of  S. 

For  safe  stresses,  which  must  lie  well  within  the  elastic  limit, 
the  unit  stress  is  proportional  to  the  stretch,  and  plane  cross- 
sections  of  the  bar  before  the  force  is  applied  are  assumed  to  re- 
main plane  after  the  bar  is  stretched.  It  is  impossible  to  detect 
experimentally  that  this  assumption  is  not  true.  Were  the  plane 
sections  to  become  even  slightly  warped,  the  cumulative  warpings 
of  successive  sections  in  a  long  bar  ought  to  become  apparent  to 
the  eye.  No  reference  is  intended  here  to  local  distortion  preceding 
failure. 

If  the  stress  on  any  section  is  not  uniform  and  the  successive 
sections  remain  plane,  they  must  be  a  little  inclined  to  one  another. 

The  stress  on  any  cross-section  5 
must  therefore  vary  uniformly  in  the 
direction  of  the  deviation  of  the 
action  line  of  Pf  from  the  centre, 
(Fig.  56),  and  be  constant  on  lines 
at  right  angles  to  that  deviation. 
The  force,  P',  acting  at  a  distance, 
y0,  from  the  axis  of  the  tie  is  equivalent 
to  the  same  force,  P',  at  the  centre 
and  a  moment,  Pfy$.  The  stress 
on  each  particle  of  any  section  may 

therefore  be  divided  into  two  parts,  that  due  to  direct  tension  and 
that  due  to  the  moment  or  to  bending.  The  former  is  uniform 
across  the  section  and  is 


The  latter  is  zero  at  the  centre  of  gravity  of  the  section  and  a 
maximum  at  the  edge  toward  which  the  load  deviates,  where  it  is 


PIECES   UNDER   TENSION.  129 

Myi 


Note  that  ft  is  the  mean  stress,  which  is  always  the  existing  stress 
at  the  centre  of  gravity  of  the  cross-section. 

Example.  —  A  square  bar  i  in.  in  section  carries  6,000  Ib.  ten- 
sion. The  centre  of  the  eye  at  the  end  is  i  in.  out  of  line.  Then 
/=  6,000(1  +  J-J-  12)  =  15,000  Ib.  per  sq.  in.,  2^  times  the  mean  and 
probably  the  intended  stress. 

A  bar  which  is  not  perfectly  straight  before  tension  is  applied 
to  it  tends  to  straighten  itself  under  a  pull,  but  the  stress  will 
not  become  uniform  on  a  cross-section.  The  bar  is  weaker 
in  the  ratio  of  }t  to  /,  as  it  might  carry  }S  if  the  force  were  cen-% 
tral,  but  now  can  safely  carry  only  jtS*  If  a  thrust  is  applied 
to  a  bent  bar,  there  is  a  tendency  to  increased  deviation  from  a 
straight  line,  and  to  an  increase  in  the  variation  of  stress. 

It  is  seen  from  the  example  above  that  a  small  deviation  y0 
will  have  a  decided  effect  in  increasing  /  for  a  given  P',  or  in 
diminishing  the  allowable  load  for  a  given  unit  stress.  Herein 
may  be  the  explanation  of  some  considerable  variations  of  the 
strength  of  apparently  similar  pieces  under  test;  and,  on  account 
of  such  effect,  added  to  other  reasons,  allowable  working  stresses 
may  well  be  and  are  reduced  below  what  otherwise  might  be 
used. 

115.  Curved  Piece  in  Bending.  —  Fig.  5yA  shows  a  small  length 
of  a  piece  whose  axis  is  a  plane  curve  and  whose  section  is  sym- 
metrical about  the  plane  of  curvature.  An  eccentrically  applied 
force  P,  which  is  called  +  if  compressive  and  --  if  tensile,  acts 
upon  the  section,  being  replaced  by  the  same  force  at  the  centre 
of  gravity  and  a  moment  M.  Because  of  the  bending  moment 
the  radial  plane  AB  will  assume  some  such  position  as  A'B',  if 
sections  plane  before  flexure  are  assumed  to  remain  plane  after 
flexure  as  with  straight  beams,  and  the  deformation  of  any  fibre 
will  be  proportional  to  its  distance  from  the  neutral  axis;  but  as  the 
original  length  of  all  fibres  was  not  the  same,  the  unit  deformation 
of  a  fibre  will  not  be  proportional  to  its  distance  from  the  neutral 
axis  and  hence  the  unit  stresses  on  the  section  will  not  vary  uni- 
formly. 


130  STRUCTURAL  MECHANICS. 

Let  the  length  of  an  unstressed  fibre  distant  y  from  the  gravity 
axis  be  /  and  let  its  change  of  length  under  stress  be  dl.  Then 
dl=dlQ+ydd  and 

_dl  _dl0  10     ydd  6 

="+~ 


From  the  figure  jf  =  ^»    6  =  ~p+y'    and    l^^0' 


hence 


If  p  is  the  unit  stress  at  a  distance  y  from  the  gravity  axis, 
p=EX,  and  that  the  internal  stresses  on  the  section  may  balance 
the  external  forces 

P  =  fpdS    and    M=fpdS-y,     ....     (2) 

where  zdy  is  represented  by  dS.     Multiply  (i)  by  E  and  substitute 
in  (2),  representing  dO/0  by  aj. 


Let 


J 

/Y      *\  /- 

J  \    p+y/          J 


p+y 
Then 


since 


fydS=o. 


M  P_  i    M\ 

°'~ESkp'  °~ES 


Substitute  these  values  in  (i)  and  multiply  by  E: 

P     M      M      y 


in  which 

k=~~S 
When  p  is  large  compared  with  the  depth  of  the  cross-section 


PIECES    UNDER    TENSION. 


>  5_  fy^L-L  f  y2ds    T  /* 

•/  i^+y   /°*>^  i+y/p   pJ 


.      I          ,     I 

nearly  =-,    or    £  =  -0? 


10+?     ^ 

P     My 

nearly;  andif/?=oo,   (3)  becomes  p  =  -^  +  ~j—,    the    stress   in   a 

straight  beam. 

The  value  of  k  can  be  determined  by  integration  for  simple 
forms  of  cross-section.      If  h  is  the  depth  of  the  piece,  for  the 

P,    p+w   i/hy  ,  I/AV.  i//A6 

+  •  •  •  § 


Rectangle,  k  =  —  i  +  y-  log 
Ellipse  and  Circle, 

2O/2O  \  /  2  f. 


FIG.  57. 

For  irregular  forms  &  can  be  determined  graphically  as  shown 
in  Fig.  576.  The  construction  follows  directly  from  the  general 
expression  for  k.  GP  is  drawn  Darallel  to  CD,  and  FH  to  CE. 
(Area  GP'A-Area  BPG)^-area  BDEA  =  £.  Hence  for  all  similar 
sections  having  the  centre  of  curvature  at  a  similar  distance  k  is 
the  same. 

116.  Hooks. — The  load  on  a  hook,  as  commonly  made,  is 
applied  through  the  centre  of  curvature,  whence  M  =  Pp.  If  this 
value  is  substituted  in  (3)  and  P  is  made  negative  because  the 
force  is  tensile,  and  if  p  and  y  are  replaced  by  /  and  —  yit  the 
capacity  of  the  hook  is  found  to  be 

PJ^s 


in  which  a  =  p  —  ^!=the  radius  of  the  hook  opening.     The  dis- 
tribution of  stress  on  the  section  is  as  shown. 


13  2  STRUCTURAL  MECHANICS. 

Example.  —  Fig.  570  shows  the  usual  proportions  of  crane-hook 
sections.  The  area  is  o.426h2,  y\  is  0.435^,  and  the  round  from  which 
it  can  be  forged  is  d=o.f]$h. 

Fora-±-h=         0.4  0.6  0.8  i.o 

k=         0.120         0-075         o-05S         0.040 

117.  Combined  Tie  and  Beam.  —  If  to  a  tension  member 
transverse  forces  are  applied,  or  if  it  is  horizontal  and  its  weight 
is  of  importance,  the  unit  tensile  stress  on  the  convex  edge,  due 
to  the  maximum  bending  moment,  must  be  added  to  the  unit 
stress  at  that  point  due  to  the  direct  pull.  As  in  §  114, 


and 


But  /  =  /&  +  /*   must   not   exceed   the  safe   unit   tension,   and 
the  needed  section  is,  since  I=Sr2, 


In  this  case  the  sections  may  vary,  since  the  external  bending 
moment  M  varies  from  point  to  point. 

If  the  piece  is  rectangular  in  section,  as  with  timber,  the 
formula  may  be  written 

f    6M    P  i  /6M 

=+       or    b 


In  practical  calculation  of  such  a  rectangular  section,  if  h  is 
assumed,  it  is  sufficient  to  compute  the  breadth  to  carry  M  and 
add  enough  breadth  to  carry  P,  when  the  combined  section 
will  have  exactly  /  at  the  edge. 

Example.  —  A  rectangular  wooden  beam  of  12  ft.  span  carries  a 
single  weight  of  3,000  Ib.  at  the  quarter  span,  and,  as  part  of  a  truss, 
resists  a  pull  of  20,000  Ib.  If  /=i,ooo  Ib.,  what  should  be  the 

section    under    the    weight?       ^max.=^~  —  =  81,000    in.-lb. 

81,000-6  20,000 

—  -  -  =  bh?=4&6.     If  h=i2,  6=3.37.     Also,  —  —  -  =1.67.    En- 


PIECES    UNDER   TENSION.  133 

tire  breadth  =  3.3 7 +  1.67  =  5.04.     Section=5Xi2  in.     The  same  result 
is  obtained  by  the  formula 


/6- 8 1, OOP 


12- 1,000  \         12 


+  20,000 


If  the  tensile  stress  in  a  combined  tie  and  beam  is  small  as 
compared  with  the  stress  due  to  flexure  the  above  solution  is 
accurate  enough,  but  when  the  tensile  stress  is  large  the  error 
is  considerable.  The  transverse  forces  acting  on  the  beam 
cause  it  to  deflect  and  consequently  the  line  in  which  the  direct 
pull  acts  does  not  pass  through  the  centre  of  gravity  of  each 
section  of  the  beam  and  a  bending  moment  is  produced  which 
partially  counteracts  the  bending  moment  due  to  the  transverse 
forces.  The  bending  moment  at  any  section  then  is 


in  which  v  is  the  deflection  of  the  tie-beam  at  the  section  where 
}b  is  found.  By  referring  to  Chapter  VI  it  is  seen  that  the  deflec- 
tion of  a  beam  may  be  written 


where  k  is  a  constant  depending  upon  the  way  in  which  the  beam 
is  loaded.  If  it  is  assumed  that  v  is  proportional  to  /&  in  the  case 
under  consideration  as  well  as  in  the  case  of  a  beam  not  subjected 
to  tension,  there  results 


For  a  beam  supported  at  both  ends  and  uniformly  loaded  &  =  T5¥; 
carrying  a  single  load  at  the  centre,  k  =  -£$',  carrying  a  single 
load  at  the  quarter-point,  k=  -fa-  If  the  ends  are  free  to  turn 
and  the  transverse  load  is  uniformly  distributed  or  is  a  single 
load  near  the  centre,  k  may  be  taken  as  TV» 

An  expression  for  the  fibre  stress  in  a  horizontal  steel  eye-bar 
bending  under  its  own  weight  is  readily  derived  from  the  last 
equation.  Since  a  bar  of  steel  one  square  inch  in  area  and  one 


134  STRUCTURAL  MECHANICS. 

foot  long  weighs  3.4  Ib.  and  E  is  29,000,000,  there  results  when 
all  dimensions  are  in  inches: 

4,900,000/2- 


/h\2' 

+ 23,000,0001  y) 


Example.  —  An  eye-bar  8X1  in.,  25  ft.  long,  carries  100,000  Ib.  ten~ 
sion.  }t=i2,$oo  Ib.  per  sq.  in.  If  the  effect  of  the  deflection  is  neg- 
lected, /&=  2,390  Ib.  Taking  the  deflection  into  account  by  the  last 
formula,  /&=  1,360  Ib. 

118.  Action  Line  of  P  Moved  towards  the  Concave  Side.  —  • 
It  will  be  economical,  if  it  can  be  done,  in  a  member  having  such 
compound  action,  to  move  the  line  on  which  P  acts  towards  the 
concave  side.  If  there  are  bending  moments  of  opposite  signs 
at  different  points  of  the  length,  or  at  the  same  point  at  different 
times,  such  adjustment  cannot  be  made.  If  y$  is  made  equal 
to  ^(Mmax)  -r-Pj  one-half  of  the  bending  moment  will  be  annulled 
at  the  point  where  Mmax.  exists,  and  at  the  point  of  no  bending 
moment  from  transverse  forces  an  equal  amount  of  bending 
moment  will  be  introduced.  The  unit  stresses  on  the  extreme 
fibres  at  the  two  sections  will  be  the  same,  but  reversed  one  for 
the  other. 

Example.—  A  horizontal  bar,  6X1  in.  section  and  15  ft.  long 
has  a  tension  of  45,000  Ib.  It  carries  100  Ib.  per  foot  uniformly  dis- 

tributed.    Mmax,  =  -  —  ^  -  =  33>75°  in.-lb.      .*.  y0  may  be  made 
|  in.     Then  /  from  Mmax.  =     y  =  ±  5,625  Ib.  on  either  edge.     But 


/  from  P=i±-    =  7,5oo(i±f)  =  7,5oo±2,8i2.5.     Stress 

at  top  at  ends  and  at  bottom  at  middle  10,3  1  2\  Ib.;  at  bottom  at 
ends  and  at  top  at  middle  4,68yi  Ib. 

The  extreme  fibre  stress  from  bending  moment  of  the  load 
varies  as  the  ordinates  to  a  parabola;  that  from  Pyo  is  constant. 
A  rectangle  superimposed  on  a  parabolic  segment  will  show  the 
resultant  fibre  stress  at  each  section. 

119.  Connecting-rod.  —  If  a  bar  oscillates  laterally  rapidly, 
as  does  a  connecting-rod  on  an  engine,  or  a  parallel  rod  on  loco- 
motive drivers,  there  are  forces  developed  due  to  the  acceleration, 
and  at  certain  positions  of  the  bar  these  forces  are  transverse  and 


PIECES   UNDER   TENSION.  135 

cause  bending.  When  a  parallel  rod  is  in  its  highest  or  lowest 
position  the  centrifugal  force,  due  to  the  circular  motion  of  every 
part  of  the  rod,  is  acting  at  right  angles  to  the  bar,  which  is  then 
subjected  to  a  uniformly  distributed  transverse  load  of 

wv2    w 


in  which  ?i  is  the  number  of  revolutions  per  second,  R  the  radius 
of  the  crank,  w  the  weight  of  unit  length  of  bar  and  g  the  acceler- 
ation due  to  gravity. 

The  rotating  end  of  a  connecting-rod  at  two  points  in  its  path 
is  under  the  influence  of  a  transverse  force  whose  intensity  is 
obtained  by  the  above  formula,  but  as  there  can  be  no  transverse 
force  at  the  sliding  end  due  to  acceleration,  the  rod  as  a  whole 
may  be  considered  to  be  acted  upon  by  transverse  forces  varying 
uniformly  from  wv2-^-gr  at  one  end  to  zero  at  the  other.  The 
maximum  fibre  stress  due  to  such  loading  may  be  found  and 
added  to  the  tensile  or  compressive  stress  caused  by  the  pull  or 
thrust  along  the  rod.  An  I-shaped  section  is  suitable  for  such 
members.  Owing  to  the  rapid  variations  and  alternation  of  stress, 
the  maximum  unit  stress  should  be  small.  Mass  is  disadvan- 
tageous in  such  rods. 

120.  Tension  and  Torsion.  —  A  tension  bar  may  be  subjected 
to  torsion  when  it  is  adjusted  by  a  nut  at  the  end,  or  by  a  turn- 
buckle.  The  moment  of  torsion  will  give  rise  to  a  unit  shear 
at  the  extreme  fibre,  for  a  round  rod,  of  q\  =  T  +o.ic)6d3  by  §  82, 
or  at  the  middle  of  the  side,  for  a  square  rod,  of  qi  =  T+o.2o8h3 
by  §  84,  either  of  which,  combined  with  }  =  P-t-S,  the  tensile 
stress,  will  give  #i=J/+\/(l/2  +  tf2).  See  §  86. 


Example.  —  A  round  bar,  2  in.  diameter,  to  be  adjusted  to  a  pull  of 
10,000  Ib.  per  square  inch,  calls  for  the  application  to  the  turnbuckle 
of  200  Ib.  with  an  arm  of  30  in.,  one-half  of  which  moment  may  be  sup- 
posed to  affect  either  half  of  the  rod.  If  the  turnbuckle  is  near  one 
end,  the  shorter  piece  will  experience  the  greater  part  of  the  moment. 

*2  OOO  *  2  •  *7 

0=—  -  5—^=1,010  Ib.     The  maximum  unit  tension  on  the  outside 

22  -I3 

fibres  of  the  rod  will  be  5,ooo-f\/(5,ooo2+i,9io2)  =  io,35o  Ib. 

121.  Tension  Connections.  —  If  a  tension  member  is  spliced, 
or  is  connected  at  its  ends  to  other  members  by  rivets,  the  splice 
should  be  so  made  or  the  rivets  should  be  so  distributed  across 


136  STRUCTURAL  MECHANICS. 

the  section  as  to  secure  a  uniform  distribution  of  stress.  An 
angle-iron  used  in  tension  should  be  connected  by  both  flanges 
if  the  whole  section  is  considered  to  be  efficient.  One  or  more 
rivet-holes  must  be  deducted  in  calculating  the  effective  section, 
depending  on  the  spacing  of  the  rivets.  See  §  185.  If  the  stress 
is  not  uniformly  distributed  on  the  cross-section,  the  required 


size  will  be  found  by  §  114,  S=-ji+~ 

Transverse  bolts  and  bolt-holes  are  similar  to  rivets  and  rivet- 
holes. 

Timbers  may  be  spliced  by  clamps  with  indents,  and  by 
scarfed  joints,  in  which  cases  the  net  section  is  much  reduced; 
so  that  timber,  while  resisting  tension  well,  is  not  economical 
for  ties,  on  account  of  the  great  waste  by  cutting  away.  How- 
ever, where  the  tie  serves  also  as  a  beam,  timber  may  be  very 
suitable. 

122.  Screw-threads  and  Nuts.  —  If  a  metal  tie  is  secured  by 
screw-threads  and  nuts,  the  section  at  the  bottom  of  the  thread 
should  be  some  15  per  cent,  larger  than  the  given  tension  would 
require,  to  allow  for  the  local  weakening  caused  by  cutting  the 
threads.  Bars  are  often  upset  or  enlarged  at  the  thread  to  give 
the  necessary  net  section  and  thus  save  the  material  which  would 
be  needed  for  an  increase  of  diameter  throughout  the  length 
of  the  rod. 

To  avoid  stripping  the  thread,  the  cylindrical  surface,  whose 
area  is  the  circumference  at  the  bottom  of  the  thread  multiplied 
by  the  effective  thickness  of  the  nut,  should  be,  when  multiplied 
by  the  safe  unit  shear,  at  least  equal  to  the  net  cross-section  of 
the  rod  multiplied  by  the  safe  unit  tension.  27iRi-t-qi=7zRi2fj 
or  }Ri  =  2qit.  As  q\  is  usually  taken  less  than  /;  as  with  a 
square  thread  only  half  of  the  thickness  is  effective,  and  with 
a  standard  V  thread  quite  a  portion  of  the  thickness  must  be 
deducted,  nuts  are  usually  given  a  thickness  nearly  or  quite 
equal  to  the  net  diameter  of  the  rod.  Heads  of  bolts  may  be 
materially  thinner. 


CHAPTER  IX. 
COMPRESSION  PIECES— COLUMNS,  POSTS,  AND  STRUTS. 

123.  Blocks  in  Compression. — If  the  height  of  the  piece  is 
quite  small  as  compared  with  either  of  its  transverse  dimensions, 
and  the  load  upon  it  is  centrally  imposed,  the  load  or  force  P  may 
reasonably  be  considered  as  uniformly  distributed  over  the  cross- 
section  S,  and  the  unit  stress  /  upon  each  square  inch  of  section 
will  be  given  by  the  formula 

as  is  the  case  with  any  tension  member  when  the  force  is  centrally 
applied. 

124.  Load  not  Central. — So  also    when    the    action    line  of 
the  resultant  load  cannot  be  considered  as  central,  but  deviates 
from  the  axis  of  the  piece  a  distance  yo,  the  force  P  can  be  replaced 
by  the  same  force  acting  in  the  axis  and  a  couple  or  moment  Pyo, 
which   moment   must   be  resisted   at   every   cross-section    by   a 
uniformly  varying  stress,   forming  a  resisting  moment   exactly 
Ike  that  found  at  a  section  of  a  beam.     Compare  Fig.  56,  and 
change  tension  to  compression. 

If  }c  is  the  uniform  unit  stress  due  to  a  central  load  of  P,  and 
if  ji  is  the  unit  stress  on  the  extreme  fibre  lying  in  the  direction 
in  which  y$  is  measured,  the  latter  stress  being  due  to  the  moment 

Py* 

p 


137 


138  STRUCTURAL  MECHANICS. 

The  greatest  unit  stress  on  the  section  is 


The  load  that  such  a  piece  will  carry  is 


By  comparison  with  the  formula  of  the  preceding  section 
it  will  be  seen  that  the  piece,  when  the  load  is  eccentric,  is  weaker 
in  the  ratio 


The  values  of  yiS+I  or  yi+r2  are  given  below  for  some 
of  the  common  sections  of  columns,  y\  being  measured  in  the 
direction  h. 

bh3  16 

Rectangle $  h               bh                        — 

12  ft 

h*  6 

Square \h                h*                         -=- 

12  h 

7T(/4  8 

Circle %d              fad2                       — 

64  a 

bh*-Vh*  ..     ,,7,          6h(bh-b'h') 

Hollow  rectangle. ..   -  \h           bh—r1-' 
12 


Hollow  circle. 


64 

125.  The  Middle  Third.  —  The  mean  or  average  unit  stress 
is  always  found  at  the  centre  of  gravity  of  the  cross -section. 
On  sections  symmetrical  about  the  neutral  axis  the  fibre  stress 
can  be  zero  at  one  edge  only  when  the  fibre  stress  on  the  oppo- 
site edge  is  twice  the  mean,  that  is,  when 


or        ;yo  =  -. 


COMPRESSION  PIECES.  139 

This  condition  is  satisfied  for  a  rectangle  when  yo  =  JA.  Hence  for 
a  rectangular  section  in  masonry  the  centre  of  pressure  must 
not  deviate  from  the  centre  of  figure  more  than  one-sixth  of 
the  breadth  in  either  direction,  if  the  unit  stress  at  the  more 
remote  edge  is  not  to  be  allowed  to  become  zero  or  tension.  As 
masonry  joints  are  supposed  in  many  cases  not  to  be  subjected 
to  tension  in  any  part,  the  above  statement  is  equivalent  to  say- 
ing that  the  centre  of  pressure  or  line  of  the  resultant  thrust 
must  always  lie  within  the  middle  third  of  any  joint. 

Likewise,  for  two  cylindrical  blocks  in  end  contact,  the  centre 
of  pressure  should  fall  within  the  middle  jourth  of  the  diameter 
if  the  pressure  is  assumed  to  be  uniformly  varying  and  it  is  not 
permissible  to  have  the  joint  tend  either  to  open  or  to  carry  ten- 
sion at  the  farther  edge. 

The  unit  pressure  at  the  most  pressed  edge  of  a  rectangle 
can  be  found  for  any  deviation  y$. 

ist.  When  the  stress  is  over  the  whole  joint,  as  before, 


2d.  When  compression  alone  is  possible,  and  only  a  part 
of  the  surface  of  the  joint  is  under  stress.  The  distance  from 
the  most  pressed  edge  to  the  action  line  of  P  is  ^k—y0.  The 
entire  pressed  area  =  3(%h—yo)b,  since  the  ordinates  representing 
stresses  make'  a  wedge  whose  length  along  y  is  three  times  the 
distance  of  P  from  the  most  pressed  edge. 


If  the  case  is  that  of  a  wall,  and  P  is  the  resultant  force  per 
unit  of  length,  6  =  1. 

As  yo  increases,  /  increases,  until  finally  the  stone  crushes 
at  the  edge  of  the  joint,  or  shears  on  an  oblique  plane  as  described 
in  §  23.  Sometimes  the  pressure  is  not  well  distributed,  from 
poor  bedding  of  the  stones,  and  spalls,  or  chips,  under  the  action 
of  the  shearing  above  referred  to,  may  break  off  along  the  edge, 
without  failure  being  imminent,  since  when  the  high  spots  break 
off  others  come  into  bearing. 

P  can  never  traverse  one  edge  of  the  joint,  if  tension  is  not 


140  STRUCTURAL  MECHANICS. 

possible  at  the  other  edge,  as  the  unit  stress  then  becomes  infinite. 
Some  writers  commit  an  error  in  determining  the  thickness  of 
a  wall  by  equating  the  moment  of  the  overturning  force  about 
the  front  edge  or  toe  with  the  moment  of  the  weight  of  the  wall 
about  the  same  point.  This  process  is  equivalent  to  making 
the  action  line  of  P  traverse  that  point.  The  centre  of  moments 
should  be  taken  either  at  the  outer  edge  of  the  middle  third  of 
the  joint,  when  pressure  is  desired  over  the  whole  joint,  or  about 
a  point  at  such  a  distance  %h—yo  from  the  front  as  will  give 
maximum  safe  pressure  at  the  front  edge.  A  uniformly  vary- 
ing stress  extending  over  three  times  that  distance  will  equal  P, 
as  lately  stated.  A  portion  of  the  joint  at  the  rear  will  then 
tend  to  open. 

Examples.  —  A  short,  hollow,  cylindrical  column,  12  in.  external 
diameter,  10  in.  internal  diameter,  supports  a  beam  which  crosses  the 
column  2  in.  from  its  centre. 

8-12          96  .    P/T,2'96\      ,,          ^ 

=  ==~ 


or  the  stress  at  either  edge  will  be  80%  greater  and  less  than  the  mean 
stress. 

A  joint,  10  ft.  broad,  of  a  retaining-wall  is  cut  at  a  point  3  ft.  9  in. 
from  the  front  edge  by  the  line  of  the  resultant  thrust  above  that  joint. 
If  this  thrust  per  foot  of  length  of  the  wall  is  28,000  lb.,  the  pressure  per 

square  foot  at  the  front  edge  will  be  -     —  (i  +  iJ-T6o-)=  i}-  2,800=4,900 
lb.     At  the  rear  it  will  be  700  lb.  per  sq.  ft. 

126.  Resistance  of  Columns.  —  A  column,  strut,  or  other 
piece,  subjected  to  longitudinal  pressure,  is  shortened  by  the 
compression.  As  perfect  homogeneousness  does  not  exist  in 
any  material,  the  longitudinal  elements  will  yield  in  different 
amounts,  so  that  there  is  apt  to  be  a  slight,  an  imperceptible 
tendency  to  curvature  of  the  strut.  Hence  the  action  line  of 
the  applied  load  may  not  traverse  the  centres  of  all  cross  -sec- 
tions of  the  piece.  The  product  of  the  applied  force  into  the 
perpendicular  distance  of  its  line  of  action  from  the  centre  of 
any  cross-section  will  be  a  bending  moment,  which  must  develop 
a  resisting  moment  at  the  cross-section,  resulting  in  a  varying 
stress,  as  in  §  124.  This  curvature  under  longitudinal  pressure 


COLUMNS,  POSTS,  AND   STRUTS.  141 

can  be  readily  obtained  with  test  specimens  of  most  materials, 
even  with  some  samples  of  cast  iron,  and  the  form  of  the  curve 
apparently  conforms  to  the  one  to  be  deduced  by  theory.  A 
tendency  to  such  a  curve  must  therefore  exist  under  working 
stresses,  although  the  curvature  is  imperceptible,  unless  the 
column  happens  to  have  its  load  perfectly  axial,  a  contingence 
that  cannot  be  safely  relied  upon.  The  column  formula,  so 
called,  should  therefore  be  confidently  applied. 

Further,  as  such  curvature  can  be  produced  in  test  speci- 
mens not  more  than  four  or  five  diameters  long,  such  a  formula 
is  applicable  to  columns  and  struts  of  any  length.  It  is  not 
necessarily  to  be  applied,  however,  to  very  short  posts,  or  blocks, 
for  the  relation  P=jS  will  determine  their  size  with  sufficient 
exactness. 

127.  The   Yield-point   Marks  the   Column   Strength.  —  The 
influence  of   yo   in   determining  the  load    a  compression   piece 
will  carry  has  been  shown  in  §  124  to  be  very  marked.     A  column 
which  has  become  sensibly  bent  under  a  load  is  very  near  com- 
plete failure.     The  moment  of  the  load  at  the  cross-section  of 
greatest  lateral  deflection  has   then   become  so   large  that   the 
stress  on  the  extreme  fibres  passes  the  yield-point,  and  the  great 
increase  of  deformation   at  and   above  the   yield-point  at  once 
increases   the  bending  moment    greatly.     Hence   it  is  true  that 
the  yield-point  marks  nearly  the  ultimate  compressive  strength 
of  materials  when  tested  in  column  form. 

Again,  the  fact  that,  in  tests  of  large  columns,  a  very  slight 
shifting  of  the  point  of  application  of  the  load  at  either  end  has 
a  decided  influence  on  the  amount  of  weight  such  a  column 
will  carry  is  a  confirmation  of  the  statements  with  which  this 
discussion  opened.  It  also  has  a  bearing  upon  the  truth  of  the 
theoretical  deduction  as  to  the  effect  of  eccentric  loading,  as 
discussed  in  §  124,  and  to  be  applied  to  long  columns  later. 

128.  Direction     of    Flexure. — The    flexure    usually    occurs, 
unless  there  is  some  defect  or  weakness,  in  a  direction  parallel 
to  the  least  transverse  dimension  of  the  strut,  i.e.,  perpendicular 
to  that  axis  in  the  cross-section  which  offers  the  least  resisting 
moment.     By  the  application  of  longitudinal  pressure  to  a  slender 


142  STRUCTURAL  MECHANICS. 

rod  its  flexure  may  be  made  very  apparent.     The  form  of  the 
column  formula  ought  to  resemble  that  of  §  124, 


129.  The  Ideal  Column.  Euler's  Formula.  —  Assume  the 
column  to  be  perfectly  straight  and  homogeneous  and  the  load 
to  be  applied  axially.  Under  such  ideal  conditions  the  load  can 
cause  no  flexure,  but  if  a  small  deflection  is  given  the  column 
by  the  application  of  a  lateral  force,  it  is  desired  to  find  what 
load  is  just  sufficient  to  maintain  that  deflection  after  the  lateral 
force  is  removed. 

If  the  column  is  fixed  in  direction  at  its  ends,  by  its  connections 
to  other  pieces,  or  by  having  a  broad,  well-bedded  base  and  cap, 
it  will  act  in  flexure  much  as  a  beam  fixed  at  the  ends. 
A  couple  or  bending  moment,  which  may  be  repre- 
sented by  MQ,  will  thus  be  introduced  at  each  end. 
Let  P  =  applied  external  force  or  load;  -z;  =  any  deflec- 
tion ordinate,  measured  at  right  angles  to  the  action 
line  of  P,  from  the  original  axis  of  the  column  to  any 
j    "\        point  in  the  axis  when  bent;    x  =  distance  from  one 
end  along  the  original  axis  to  any  ordinate  v;  1  =  length 
D  1      of  column. 

The  combination  of  the  moment  MQ  at  the  end 
of  the  column  with  the  force  P  has  the  effect,  as  shown 
in  mechanics,  of  shifting  the  force  P  laterally  a  distance 
MQ  -±-P=VQ-,  hence  the  action  line  of  P  is  now  parallel 
to  the  original  axis,  at  a  distance  VQ  from  it,  or  in  the 
line  F  E  of  Fig.  58.     The  ordinate  to  the  points   of 
pj  |         contraflexure   is    therefore   VQ.    This    action   can   be 
!    J          more  fully  realized   by  conceiving  that   the  bearing 
surface  at  A  is  removed,  and  that  P  acts  at  such  a  point 
on  a  horizontal  lever  as  to  keep  the  tangent  to  the  curve  at  A 
strictly  vertical. 

As  v  is  measured  from  the  original  axis,  the  bending  moment 
at  any  section  is  M  =  P(v0—v),  which  will  change  sign  when  the 
second  term  is  larger  than  the  first. 


COLUMNS,   POSTS,   AND  STRUTS.  143 

If  the  flexure  is  very  slight,  an  equation  similar  to  that  used 
with  beams  may  be  written 

d2v 


dv 
Multiply  by  -j-  and  integrate. 


As  -j-  =  o  when  v  =  o,  C=o.    When  T-=O  at  D,  the  middle 
dx  doc 

of  the  length,  vmax,  =  2V0,  or  double  the  ordinate  at  the  point  of 
contraflexure,  F.  Extracting  the  square  root  of  the  above  equation 
and  separating  the  variables  gives 

~EJ        dv 


\//2V0V—V2' 

m 


As  v=o  when  #  =  o,  C'  =  o. 


V=VQ  versm 
As  i  —cos  6  =  2  sin2J#, 

V  =  2VQ 


/      \~P\         /  (      \P\\ 

m  (  ^\|^j  )  =  ^ol  i  -cos  \x\j^j))' 


If,  in  this  equation,  x  =  ^l  a  value  of  iVz*.  is  obtained  to  be 
equated  with  the  previous  value,  2i>o- 


'Umax.  =  2V0  =  2V0  Sm2  I  t^Nj^T 


P 

sm 


144  STRUCTURAL  MECHANICS. 

which  is  commonly  known  as  Eulefs  formula.  P  is  the  ultimate 
strength  of  the  column  and  is  independent  of  the  deflection;  that 
is,  within  certain  limits  P  is  the  same,  no  matter  what  small 
deflection  is  given  the  column.  Under  a  load  less  than  P  the 
column  would  straighten  ;  under  a  greater  load  it  would  fail. 
The  equation  of  the  curve  of  the  column  is 


2 


=  2V0  sn 


To  find  the  points  of  contraflexure,  make  V=VQ. 

sin  (y#)  =\/i  =  sin  45°  =  sin—. 

oc 
•*•       =  or        x  =     from  either  end. 


Hence  the  curve  is  made  of  four  equal  portions,  A  F,  F  D, 
D  E,  and  E  H. 

A  column  hinged,  pin-ended,  or  free  to  turn  at  its  ends,  and 
of  length  represented  by  EF  =  J/,  will  have  the  same  portion 
of  stress  at  D  that  is  due  to  bending  as  does  a  column  of  length 
AH=/,  which  is  fixed  at  its  ends. 

If,  in  actual  cases,  F  is  considered  to  be  practically  in  the 
same  position  horizontally  as  before  loading,  it  may  be  said 
that  a  column  fixed  at  one  end  and  hinged  at  the  other,  of  length 
F  H  =  |/,  will  also  have  the  same  portion  of  stress  arising  from 
bending.  The  maximum  deflection  will  then  occur  at  one-third 
of  its  length  from  the  hinged  end.  This  result  has  been  verified 
by  direct  experiment  on  a  full-sized  steel  bridge  member. 

130.  Rankine's  Column  Formula.  —  Euler's  formula  can  be 
put  in  the  form 

P 


P  I 

This  curve,  when  plotted  with  -~-  and   —  as  variables,  has 

ij  Y 

the  form  of  the  curve  B  C,  Fig.  59,  where  the  greatest  mean  unit 


COLUMNS,  POSTS,  AND  STRUTS. 


145 


stress  which  any  ideal  column  of  a  given  ratio  of  /  to  r  can  endure 
is  given  by  the  ordinate  to  the  curve.  But  as  the  yield-point 
marks  the  ultimate  strength  of  metal  in  compression,  for  short 
columns  the  line  A  B,  whose  ordinate  is  the  yield-point,  takes 
the  place  of  the  curve. 

For  long,  slender  columns  (when  l-^-r  exceeds  200,  say) 
Euler's  formula  gives  results  close  to  the  ultimate  strength  found 
by  loading  actual  columns  to  destruction;  but  for  shorter  columns 
the  formula  gives  results  much  too  large.  It  must  be  remembered 
that  actual  columns  do  not  satisfy  the  conditions  from  which 
Euler's  formula  was  derived.  No  real  column  is  perfectly  straight 


—  30000- 


1,144,000,000 


—20000- 


-10000- 


40,000 

— 


28,600  r  I 


100 


200 


300 


400 


Fig.  59 


and  homogeneous,  nor  can  the  load  be  applied  exactly  along 
the  axis,  and  as  no  two  columns  are  exactly  alike  no  theoretical 
formula  can  exactly  fit  all  cases.  For  these  reasons  Euler's 
formula  is  never  used  in  designing. 

When  the  results   of  actual  column  tests  are  plotted,   they 
follow  in  a  general  way  some  such  curve  as  ADC.     Rankings 


formula,  -~-  = 


P 


,  is  the  equation  of   such  a  curve,  in  which 


p  is  the  yield-point  of  the  material  and  a  is  a  numerical  coefficient. 
By  modifying  the  value  of  a  the  curve  can  be  made  to  agree 
fairly  well  with  the  results  of  tests.  For  a  short  block  when 


146  STRUCTURAL  MECHANICS. 

l+r  approaches  zero  the  formula  becomes  P+S  =  p.  If  a  is 
made  equal  to  p  +4~2E,  the  curve  of  Rankine's  formula  approaches 
Euler's  for  large  values  of  l+r,  where  Euler's  agrees  well  with 
the  results  of  tests.  This  value  of  a  has  been  used  by  some 
authorities  and  is  believed  to  give  conservative  values.  The 
curve  of  Fig.  59  was  so  plotted.  However,  the  value  of  any 
column  formula  depends  upon  its  agreement  with  the  results 
of  tests,  and  in  practice  a  is  generally  an  empirical  factor. 

In  order  that  P  of  Rankine's  formula  may  be  the  load  which 
an  actual  column  can  safely  carry  instead  of  its  ultimate  strength, 
the  greatest  allowable  unit  stress  on  the  material,  /,  is  to  be  sub- 
stituted for  the  yield-point,  p.  The  formula  then  becomes 

P         } 


5  P' 

i+<75 

It  is  to  be  regarded  as  giving  the  mean  unit  stress  on  any  cross- 
section  when  the  stress  on  the  extreme  fibre  at  the  point  of  greatest 
deflection  is  /. 

In  designing  columns  the  load,  P,  and  the  length,  /,  are  usu- 
ally known.  The  form  of  cross-section  is  then  assumed  which 
fixes  S  and  r.  If  the  assumed  values  satisfy  the  equation  the 
section  chosen  is  satisfactory;  if  not,  another  trial  must  be  made. 
The  radius  of  gyration,  r,  to  be  used  is  the  one  perpendicular 
to  the  neutral  plane,  that  is,  it  is  measured  in  the  direction  in 
which  the  column  is  likely  to  deflect.  Unless  the  column  is 
restrained  in  some  way  it  will,  of  course,  deflect  in  the  direction 
of  the  least  radius  of  gyration  of  the  cross- section. 

Example. — What  load  will  a  i2-in.  3iJ-lb.  I  beam  10  ft.  long  carry 

as  a  column  if  the  mean  unit  stress  is  12,000-^-  ( i  H —       — ~ )  ?  r  about 

\      36,ooorv 


/       14,400 
axis  through  web=i.oi  in.       S=  9.  26  sq.  in.        -%=  -  =  14,100. 

=8,620  Ib.     P=8,62oX9.26=  79,800  Ib. 


.. 
36,000  1.392 

131.  Multipliers  of  a.  —  As  seen  above,  2/  should  theoretically 
be  substituted  for  /  when  the  column  is  hinged  or  free  to  turn 


COLUMNS,  POSTS,  AND  STRUTS.  147 

at  its  ends,  in  order  to  obtain  the  equivalent  length  of  a  column 
which  is  fixed  at  the  ends;  and  for  a  column  fixed  at  one  end 
and  hinged  at  the  other  f/  should  be  substituted  for  /  for  the 
same  reason.  Or,  more  conveniently,  a  may  be  used  for  a  column 
fixed  at  the  ends  and  of  length  /;  40  for  a  column  hinged  at 
both  ends  and  of  length  /;  and  -l-f-a  for  a  column  hinged  at  one 
end  and  fixed  at  the  other,  length  /.  Actual  tests,  carried,  how- 
ever, to  the  extreme  of  bending  or  crippling,  appear  to  show 
that  a  column  bearing  on  a  pin  at  each  end  is  not  hinged  or  per- 
fectly free  to  turn;  hence  the  multipliers  of  a  more  commonly 
used,  instead  of  i,  *-/-,  and  4,  are  i,  f,  and  2. 

132.  Pin   Friction. — Some   regard    columns    as    neither   per- 
fectly fixed  nor  perfectly  hinged,  and  use  but  one  value  of  a 
for  all,  which  might  perhaps  then  be  taken  as  a  mean  value. 
The  moment  of  friction  on  a  pin  is  considerable.     If  P  is  the 
load  on  a  post  or  strut,  d  the  diameter  of  the  pin,  and  tan  </>  the 
coefficient  of  friction  of  the  post  on  the  pin,  the  moment  of  fric- 
tion at  the  pin  will  be  P-Jd-tan  <£;   and  this  moment,  if  greater 
than  MQ  =  PVQ,  will  keep  the  post  restrained  at  the  end,  so  that 
the  tangent  there  to  the  curve  remains  in  its  original  direction. 
As  P  and  VQ  increase,  MO  will  become  the  greater  when  VQ  exceeds 
\d  tan  <£;   the  column  will  then  be  imperfectly  restrained  at  its 
ends,  and  the  inclination  will  change.     As  the  friction  of  motion 
is  less  than  that  of  rest,  such  movement  when  started  may  be 
rapid.     Some  tested  columns,  showing  at  first  the  curve  of  Fig.  58 
known  as  triple  flexure,  have  suddenly  sprung  into  a  single  curve 
and  at  once  offered  less  resistance. 

It  may  be  doubted  whether  ordinary  columns  under  working 
loads  ever  develop  a  value  of  v0  sufficient  to  overcome  the  pin 
friction,  unless  the  column  is  very  slender  and  the  diameter 
of  the  pin  small.  The  custom  is  quite  general  of  using  the  same 
column  formula  for  struts  with  fixed  and  with  hinged  ends. 

133.  Straight-line   Column   Formula. — In   engineering  struc- 
tures columns  which  have  a  greater  ratio  of  length  to  radius 
of  gyration  than  100  or  150  are  very  rarely  used.     Within  this 
limit  a  straight  line  can  be  drawn  which  will  represent  the  aver- 
age results  of  a  series  of  column  tests,  plotted  as  described  in 
§  130,  as  well  as  Rankine's  formula  does.     As  the  equation  of  a 
straight  line  is  easier  to  solve  than  the  equation  of  a  curve,  the 


148 


STRUCTURAL  MECHANICS. 


straight-line  formula  is  much  used.     For  working  values  it  has 
the  form 

p  ,   i 


in  which  /  is  the  greatest  allowable  unit  stress  on  the  material 
and  c  is  a  numerical  coefficient  determined  empirically.  In  this 
country  both  Rankine's  and  the  straight-line  formulas  are  exten- 
sively used  in  designing  columns^  Examples  of  each  will  be 
found  in  Chapter  X. 

134.  Swelled  Columns.  —  Some  posts  and  struts,  especially 
such  as  are  built  up  of  angles  connected  with  lacing,  are  swelled 
or  made  of  greater  depth  in  the  middle.  If  the  strut  is  perfectly 
free  to  turn  at  the  ends,  such  increase  in  the  value  of  r2  may  be 
quite  effective,  and  r2  for  the  middle  section  may  be  used  in 
determining  the  value  of  P,  provided  the  latter  does  not  too 
closely  approach  the  uniform  compression  value  at  the  narrower 
ends.  But  if  the  strut  is  fixed  at  the  ends,  or  is  attached  by  a  pin 
whose  diameter  is  large  enough  to  make  a  considerable  friction- 
moment,  such  enlargement  at  the  middle  is  useless  ;  for  an  equally 
large  value  of  r2  ought  to  be  found  at  the  ends  also.  Hence 
swelled  columns  and  struts  are  but  little  used. 


135.  Column  Eccentrically  Loaded. 

— When  the  load  is  applied  eccen- 
trically to  a  long  column,  the  maxi- 
mum unit  stress  found  in  the  extreme 
fibre  on  the  concave  side  must  be  due 
to  three  combined  effects : 

ist.  The  stress  due  to  the  load  P, 
or  po  =  P+S.  (Fig.  60.) 

2d.  The  stress  due  to  the  resisting 
moment  set  up  by  Pyo. 

3d.  The  stress  due  to  the  resisting 
moment  set  up  by 


Fig.  60 


From  §  124, 


From  §  130, 


12 


COLUMNS,  POSTS,  AND  STRUTS.  149 

If  then  the  column  is  long  and  the  line  of  action  of  P  deviates 
from  the  original  axis  of  the  column  by  a  distance  ;yo,  the  three 
expressions,  po,  poyoyi+r2,  and  p0al2  +r2  should  be  added,  giving 

•  L=  f 

"  S  I2 


This  formula  may  be  put  in  a  form  more  convenient  for 
designing  : 

P        Py0yi 


The  first  term  gives  the  area  of  cross-section  necessary  to 
resist  the  direct  thrust  and  the  second  term  the  area  to  resist  the 
bending  moment  due  to  eccentric  loading. 

Since  y0  will  determine  the  direction  of  flexure,  r  must  be 
taken  in  this  case  in  the  direction  yo',  that  is,  the  moment  of 
inertia  and  r  must  be  obtained  about  the  axis  through  the  centre 
of  gravity  and  lying  in  the  plane  of  the  section,  perpendicular  to 
y0.  That  the  moment  Pyo,  although  small,  has  a  decided  weaken- 
ing effect  on  a  column  is  proved  by  experiment,  and  its  unintended 
presence  may  explain  some  anomalies  in  tests. 

136.  Transverse  Force  on  a  Column.  —  The  resisting  power 
of  a  column  or  strut  to  which  a  transverse  force  is  applied  in  ad- 
dition to  the  load  in  the  direction  of  the  axis,  and  the  proper 
dimensions  of  the  strut,  are  involved  in  some  doubt.  Theo- 
retically, the  formula  is  deduced  as  follows: 

From  the  formula  for  the  resisting  moment  of  a  beam, 
M  =  }I+yij  the  stress  on  the  outer  layer  from  such  bending 
moment  is  My\  +1.  Hence  if  M  is  that  particular  value  of  the 
bending  moment  (from  the  transverse  load  or  force)  which  exists 
at  the  section  of  maximum  strut  deflection,  where  the  column  stress 
is  greatest  (that  is,  at  the  middle  for  a  column  with  hinged  ends,  but 
perhaps  at  the  ends  for  a  column  with  fixed  ends,  since  M  may 
then  be  greater  at  the  ends),  the  maximum  unit  stress  on  the 
concave  side 


STRUCTURAL  MECHANICS. 


If  the  straight-line  formula  is  used  instead  of  Rankine's,  the 
last  equation  has  the  form 


Or,  again,  it  may  be  said  that  at  the  section  in  question  P  is 
moved  laterally  by  the  moment  M  a  distance  yo  =  M  '  -r-P.  Then, 
by  §  135, 

P_        ! 

S  I2    My  i9 

i  +  a—,  +  -T—T- 
r2     Pr2 

which  reduces  to  the  forms  given  above. 

The  value  of  r2  to  be  used  in  these  formulas  is  that  for  flexure 
in  the  plane  of  M. 

Example.  —  A  deck  railway-bridge  with  2o-ft.  panels  has  ties  laid 
directly  on  the  top  chord,  thus  imposing  a  load  of  2,500  Ib.  per  foot.  If 
the  direct  thrust  is  249,000  Ib.,  what  should  be  the  size  of  chord  for  a 
working  stress  on  columns  of  10,000—  4^1-^  r  and  on  beams  of  9,000  Ib. 
per  sq.  in.  ?  Try  a  section  composed  of  two  2o-in.  65-lb.  I  beams  and 
one  24Xi-in.  cover-plate.  5=50.16  sq.  in.  The  centre  of  gravity  of 
the  section  lies  2.45  in.  above  middle  of  I  beams.  y\  to  upper  fibre  = 
8.05  in.  7=3,298  in.4  r  about  horizontal  axis=8.n  in.  M= 
J-  2,500-  20-  20-  12=  i,  500,000  in.-lb.  Working  column  stress=  10,000 
—  45X240-^8.11  =  8,670  Ib.  per  sq.  in.  249,000-^8,670=28.7  sq.  in. 

....  .  1,500,000X8.05  .  . 

required  for  column  action.     —    —5  -  Vy0      =  20.4  sq.  in.  required 

9,000X8.11X8.11 

for  beam  action.     28.7  +  20.4=49.1  sq.  in.,  total.     Assumed  section  is 
sufficient. 


COLUMNS,  POSTS,  AND  STRUTS.  151 

137.  Lacing-bars. — The  parts  of  built-up  posts  are  usually 
connected  with  lacing  straps  or  bars.  These  bars  carry  the 
shear  due  to  the  bending  moment  arising  from  the  tendency  of 
the  post  to  bend  and  should  be  able  to  stand  the  tension  and 
compression  induced  by  the  shear.  At  the  ends  and  where 
other  members  are  connected,  in  order  to  insure  a  distribution 
of  load  over  both  members,  batten  or  connection  plates  at  least 
as  long  as  the  transverse  distance  between  rivet  rows  are  used 
in  good  practice. 

Each  piece  should  be  of  equal  strength  throughout  all  its 
details.  A  post  or  strut  composed  of  two  channels  connected 
by  lacing-bars  and  tie-plates  is  proportioned  for  a  certain  load, 
the  mean  unit  stress  being  reduced  in  accordance  with  the  formula 
in  which  the  variable  is  the  ratio  of  the  length  to  the  least  radius 
of  gyration  of  the  whole  section.  In  the  lengths  between  the 
lacing-bars,  this  ratio  for  one  channel  with  its  own  least  radius 
should  not  be  greater  than  for  the  entire  post.  Nor  should  the 
flange  of  the  channel  have  any  greater  tendency  to  buckle  than 
should  one  channel  by  itself.  The  same  thing  applies  to  the 
ends  of  posts,  where  flanges  are  sometimes  cut  away  to  admit 
other  members.  Quite  a  large  bending  moment  may  be  thrown 
on  such  ends,  when  the  plane  of  a  lateral  system  of  bracing  does 
not  pass  through  the  pins  or  points  of  connection  of  the  main 
truss  system. 

The  usual  bridge  specification  reads :  Single  lacing-bars  shall 
have  a  thickness  of  not  less  than  ^V,  and  double  bars,  connected 
by  a  rivet  at  their  intersection,  of  not  less  than  -£$  of  the  distance 
between  the  rivets  connecting  them  to  the  member.  Their 
width  shall  be:  For  i5-in.  channels,  or  built  sections  with 
3^-  and  4-in.  angles,  2j  in.,  with  f-in.  rivets;  for  12-  or  lo-in. 
channels,  or  built  sections  with  3-in.  angles,  2j-in.,  with  f-in. 
rivets;  for  9-  or  8-in.  channels,  or  built  sections  with  2j-in.  angles, 
2  in.,  with  f-in.  rivets. 

All  segments  of  compression  members,  connected  by  lacing 
only,  shall  have  ties  or  batten  plates  placed  as  near  the  ends  as 
practicable.  These  plates  shall  have  a  length  of  not  less  than 
the  greatest  depth  or  width  of  the  member  and  shall  have  a  thick- 
ness of  not  less  than  -fa  of  the  distance  between  the  rivets  con- 
necting them  to  the  compression  member. 

Examples. — i.  A  single  angle-iron  6X4X|  in.  and  6  ft.  8  in.  long 
is  in  compression.  Use  r=o.g  or  obtain  it  from  §  76.  5=3.61  sq.  in. 
If  P-r- S  =  12,000  —  34/-^r,  what  will  it  carry?  32,400  Ib. 

2.  A  square  timber  post  16  ft.  long  is  expected  to  support  80,000  Ib. 


i$2  STRUCTURAL  MECHANICS. 

If  /=  1,000  and  the  subtract! ve  term  is  2/-f-r,  what  is  the  size? 

10X10  in. 

3.  What  is  the  safe  load  on  a  hollow  cylindrical  cast-iron  column 
13  ft.  6  in.  long,  6  in.  external  diameter,  and  i  in.  thick,  if  it  has  a  broad, 
flat  base,  but  is  not  restrained  at  its  upper  end?     (/=  9,000  lb.,  E= 
17,000,000,  5=15.7  sq.  in.)  124,000  lb. 

4.  If  a  short  wooden  post  12  in.  square   carries    28,800  lb.  load, 
and  the  centre  of  pressure  is  3  in.  perpendicularly  from  the  middle  of 
one  edge,  what  will  be  the  maximum  and  the  mean  unit  pressure,  and 
the  maximum  unit  tension,  if  any?  500  lb. ;  200  lb. ;  — 100  lb. 


CHAPTER  X. 
SAFE  WORKING  STRESSES. 

138.  Endurance  of  Metals  Under  Stress. — It  is  important  to 
determine  how  long  a  piece  may  be  expected  to  endure  stress 
when  constant,  when  repeated,  or  when  varied  and  perhaps 
reversed;  and  it  is  still  more  important  to  find  what  working 
stresses  may  be  allowed  upon  a  given  material  in  order  that 
rupture  by  the  stresses  may  be  postponed  indefinitely. 

The  experiments  carried  on  by  Wohler  and  Spangenberg,  and 
afterwards  continued  by  Bauschinger,  show  the  action  of  iron 
and  steel  under  repeated  stress.  Tests  were  made  on  speci- 
mens in  tension,  bending,  and  torsion.  *  A  number  of  bars  of 
wrought  iron  arid  steel  were  subjected  repeatedly  to  a  load  which 
varied  between  zero  and  an  amount  somewhat  less  than  the  ulti- 
mate strength.  When  the  bar  broke  under  the  load  a  similar 
bar  was  tested  under  a  reduced  load  and  was  found  to  resist  a 
greater  number  of  applications  before  rupture.  Finally  a  load 
was  reached  which  did  not  cause  rupture  after  many  million 
repetitions.  The  stress  caused  by  such  load  was  taken  as  the 
safe  strength  of  the  material  and  was  called  the  primitive  safe 
strength  The  following  table  is  an  illustration : 

WOHLER' s  TESTS  OF  PHCENIX  IRON  IN  TENSION. 

Stress  varying  between  Number  of  Applications. 


o 

and 

46,500 

Ib. 

per 

sq. 

in  

800 

o 

1  1 

43,000 

(  i 

<  i 

i  i 

1  1 

107,000 

0 

(  C 

39,000 

(  ( 

(  i 

i  ( 

c  i 

....       341,000 

o 

11 

35^000 

" 

li 

11 

11  

481,000 

0 

(  I 

31,000 

1  t. 

i  ( 

(  i 

1  1 

10,142,000 

*  For  a  description  of  machines  and  tests,  see  Unwin,  Testing  of  Materials  of 
Construction. 

153 


154 


STRUCTURAL  MECHANICS. 


As  a  result  of  Wohler's  and  Spangenberg's  experiments  the 
latter  states  that  alternations  of  stress  may  take  place  between 
the  following  limits  of  tensile  (— )  and  compressive  (  +  )  stress  in 
pounds  per  square  inch  with  equal  security  against  rupture: 


-15.500 
—29,000 
-43,000 

-27,000 
-47,000 
—  78,000 


PHCENIX   IRON   AXLE. 

+  15,500  Difference  31,000 

o  29,000 

-23,500  i9>5°° 

KRUPP  CAST- STEEL  AXLE. 

+  27,000     Difference  54,000 

o          "  47,000 


-34,000 


44,000 


UNHARDENED    SPRING   STEEL. 

o  Difference  48,500 

44,000 
39,000 


—  24,000 
-39,000 


-48,500 
-68,000 
-78,000 
—87,500  —58,500  29,000 

The  figures  are  approximate. 

The  results  illustrate  Wohler's  law,  that  rupture  o)  material 
may  be  caused  by  repeated  alternations  of  stress  none  of  which  attains 
the  absolute  breaking  limit.  The  greater  the  range  of  stress  the 
smaller  the  stress  which  will  cause  rupture. 

The  term  jatigue  o]  metals  is  often  applied  to  the  phenomena 
just  described. 

139.  Bauschinger's  Experiments.  —  Professor  Bauschinger, 
besides  making  tests  similar  to  Wohler's,  determined  the  effect 
which  repeated  stresses  had  upon  the  position  of  the  elastic 
limit  and  the  yield -point.  (§13,  14,  and  15.)  He  derived  the 
following  results  from  tensile  tests: 


Primitive 
Safe 
Stress,  «. 

Original 
Elastic 
Limit. 

Original 
Yield- 
point. 

Ultimate 
Strength,  t. 

\Vrought  -iron  plate          

28  ooo 

14,800 

29,700 

54  600 

BesseTier  soft  -steel  plate  

34,000 

33.QOO 

41,800 

62,000 

Iron    flat 

31  OOO 

2  C   7OO 

72  6oO 

e  7  600 

I  '-01   flat  

34,000 

32,300 

3=:.ioo 

^7,200 

Tho'Tias  steel  axle 

A  ?  OOO 

38  100 

87  ooo 

SAFE  WORKING  STRESSES  155 

From  an  inspection  of  the  table  it  is  seen  that  the  primitive 
safe  stress  on  sound  bars  of  iron  and  steel  for  tensions  alternating 
from  zero  to  that  stress  is  a  little  less  than  the  original  yield-point 
of  the  metal.  Although  the  safe  stress  was  found  to  be  greater 
than  the  original  elastic  limit,  yet  when  bars  which  remained 
unbroken  after  several  million  applications  of  stress  were  tested 
statically  it  was  found  that  the  elastic  limit  had  risen  above 
the  stress  applied.  The  inference  is  that  the  primitive  safe 
stress  is  below  the  elastic  limit  although  the  elastic  limit  may 
be  an  artificially  raised  one. 

Pulling  a  bar  with  loads  above  the  original  elastic  limit  and 
below  the  yield-point  raises  the  elastic  limit  in  tension  but  lowers 
it  in  compression  and  vice  versa.  These  artificially  produced 
elastic  limits  are  very  unstable.  When  a  bar  is  subjected  to  a 
few  alternations  of  equal  and  opposite  stresses  which  are  equal 
to  or  somewhat  exceed  the  original  elastic  limits,  the  latter  tend 
toward  fixed  positions  called  by  Bauschinger  natural  elastic 
limits.  He  advanced  the  view  that  the  original  elastic  limits 
of  rolled  steel  and  iron  are  artificially  raised  by  the  stresses  set 
up  during  manufacture.  These  natural  elastic  limits  seem  to 
correspond  with  Wohler's  range  of  stress  for  unlimited  alternating 
stresses.. 

140.  Alteration  of  Structure. — If  a  bar  has  been  subjected  to 
repetitions  of  stress  somewhat  below  the  yield-point,  any  general 
weakening  or  deterioration  in  the  quality  of  the  bar  ought  to 
show  itself  in  some  way;  but  bars  which  have  endured  millions 
of  repeated  stresses  or  bars  which  have  broken  under  repeated 
stresses  give  no  indication  of  any  weakening  when  tested  statically. 
Instead  of  being  weakened  by  repeated  stresses  well  below  the 
primitive  safe  strength,  a  bar  is  really  improved  in  tenacity. 

Bars  fractured  in  the  Wohler  machines  did  not  draw  out, 
no  matter  how  ductile  the  material  might  be,  but  broke  as  if 
the  material  were  brittle.  From  these  facts  we  are  forced  to 
conclude  that  whatever  weakness  there  may  be  is  confined  to 
the  surface  of  rupture.  Such  breaks  appear  to  be  detail  breaks. 
Continuity  is  first  destroyed  at  a  flaw  or  overstressed  spot,  and 
from  that  point  the  fracture  spreads  gradually  until  the  section 


156  STRUCTURAL  MECHANICS. 

is  so  weakened  that  the  load  is  eccentric  and  bending  stresses 
are  set  up  which  cause  the  piece  to  break  by  flexure. 

From  the  results  of  Wohler's  and  Bauschinger's  tests  it  would 
appear  that  steel  might  undergo  an  indefinite  number  of  repe- 
titions of  stress  within  the  elastic  limit  without  rupture,  or  that 
within  that  limit  the  elasticity  was  perfect.  However,  certain 
experiments  on  the  thermoelectric  and  magnetic  properties  of 
iron  under  stress  and  on  the  molecular  friction  of  torsional  pen- 
dulums give  results  inconsistent  with  the  theory  of  perfect  elas- 
ticity. Engineers  differ  more  or  less  in  their  interpretations 
of  the  experimental  results  but  agree  in  using  lower  working  stresses 
for  varying  than  for  static  loads. 

141.  Launhardt-Weyrauch  Formula. — A  number  of  formulas 
based  on  Wohler's  experiments  have  been  advanced  for  the 
determination  of  the  allowable  unit  stress  on  iron  and  steel  when 
the  range,  as  well  as  the  magnitude,  of  the  stress  is  considered. 
Launhardt  proposed  the  following  formula  for  the  breaking 
load  of  a  member  which  is  subjected  to  stresses  varying  between 
a  maximum  and  a  minimum  stress  of  the  same  kind: 


a=ul 


t  —u  min.  stressX 
u    max.  stress/ 


a = breaking  load  under  the  given  conditions. 

u  =  primitive  safe  stress. 

/  =  ultimate  strength,  static. 

v  =  greatest  stress  piece  will  bear  if  repeated  from  +v  to  —  v 
indefinitely. 

Weyrauch  extended  Launhardt's  formula  to  cover  cases  of 
alternate  tension  and  compression,  in  which  case  the  minimum 
stress  is  to  be  considered  as  negative. 


/      u  —v  mm.  stressX 
=cdi+- 
\         u 


max.  stress/' 

As  the  results  of  the  different  experiments  vary  more  or  less, 
the  different  authorities  do  not  agree  as  to  the  values  of  /,  u, 
and  v  to  be  substituted  in  the  formulas.  In  some  of  Wohler's 
experiments  u  appears  to  be  greater  than  \t  and  to  approxi- 
mate to  §/  This  value  is  assumed  by  Weyrauch  and  gives 
(t—u)+u  =  %.  Likewise  if  v  =  %u  the  coefficient  of  the  second 
formula  becomes  \  and  both  formulas  reduce  to 


SAFE  WORKING  STRESSES.  157 

min.  stressX 


2  max.  stress/ 


the  sign  of  the  second  term  changing  for  reversed  stresses.  The 
choice  of  values  of  u  and  v  to  use  in  the  equations  seems  to  have 
been  determined  largely  by  the  result  sought.  The  formulas 
are  of  rather  doubtful  value,  but  they  are  more  or  less  used  for 
determining  working  stresses  in  steel  bridges. 

The  following  relationships  of  /,  u,  and  v  are  given  by  some 
authorities,  and  they  conform  more  nearly  with  the  results  of  the 
experiments  quoted  in  this  chapter: 

Steady  load a  =  t 

Load  varying  from  o  to  u a  =  u  =  \t 

+v  to-v a=v  =  $t 

142.  Reduction  of  Unit  Stresses. — The  safe  working  stress  to 
be  used  for  any  material  will  depend  upon  several  considerations : 
Whether  the  structure  is  to  be  temporary  or  permanent;  whether 
the  load  is  stationary  or  variable  and  moving ;  if  moving,  whether 
its  application  is  accompanied  by  severe  dynamic  shock  and 
perhaps  pounding;  whether  the  load  assumed  for  calculation  is 
the  absolute  maximum;  whether  such  maximum  is  applied  rarely 
or  is  likely  to  occur  frequently;  whether  the  stresses  obtained 
are  exact  or  approximate;  whether  there  are  likely  to  be  secondary 
stresses  due  to  moments  arising  from  changes  of  the  assumed 
frame;  what  the  importance  of  the  piece  is  in  the  structure,  and 
the  possible  damage  that  might  be  caused  by  its  failure. 

The  allowable  unit  stresses  of  different  kinds,  and  for  greater 
or  less  change  of  load,  will  be  further  reduced  to  provide  against : 
Distribution  of  stress  on  any  cross-section  somewhat  different 
from  that  assumed;  variations  in  quality  of  material;  imperfec- 
tions of  workmanship,  causing  unequal  distribution  of  stress; 
scantness  of  dimensions;  corrosion,  wear,  or  other  deterioration 
from  lapse  of  time  or  neglect ;  lack  of  exactness  of  calculation. 

The  allowable  unit  stresses  so  determined  will  be  but  a  small 
fraction  of  the  ultimate  or  breaking  strength  of  the  material;  and 
it  is  evident  that  the  idea  that  it  will  require  several  times  the 
allowable  maximum  working  load  to  cause  a  structure  to  fail  is 
seriously  in  error. 


158  STRUCTURAL  MECHANICS. 

Overconfidence  of  the  in  experienced  designer  in  the  correct- 
ness of  his  design  may  be  checked  by  a  study  of  this  section. 

143.  Load  and  Impact. — The  design  should  be  completely 
carried  out,  both  in  the  principal  parts  and  in  the  details.  The 
latter  require  the  most  careful  study,  that  they  may  be  at  once 
effective,  simple,  and  practical.*  All  the  exterior  forces  which 
may  possibly  act  upon  a  structure  should  be  considered,  and  due 
provision  should  be  made  for  resisting  them.  The  static  load, 
the  live  load,  pressure  from  wind  and  snow,  vibration,  shock,  and 
centrifugal  force  should  be  provided  for,  as  should  also  deteriora- 
tion from  time,  neglect,  or  probable  abuse.  A  truss  over  a  machine- 
shop  may  at  some  time  be  used  for  supporting  a  heavy  weight  by 
block  and  tackle,  or  a  line  of  shafting  may  be  added;  a  small 
surplus  of  material  in  the  original  design  will  then  prove  of  value. 
Light,  slender  members  in  a  bridge  truss,  while  theoretically 
able  to  resist  the  load  shown  by  the  strain  sheet,  are  of  small 
value  in  time  of  accident.  The  tendency  from  year  to  year  is 
towards  heavier  construction. 

Secondary  stresses,  as  they  are  called,  are  due — first,  to  the 
moments  at  intersections  or  joints,  when  the  axes  of  the  mem- 
bers coming  together  at  a  connection  do  not  intersect  at  a  common 
point ;  and  second,  to  the  moments  set  up  at  joints  by  the  resistance 
to  rotation  experienced  by  the  several  parts  when  the  frame  or 
truss  is  deflected  by  a  moving  load.  If  symmetrical  sections 
are  used  for  the  members,  if  the  connecting  rivets  are  symmetrically 
placed,  and  if  the  axes  of  the  intersecting  members  meet  at  one 
point,  secondary  stresses  will  be  much  reduced. 

All  members  of  a  structure  should  be  of  equal  strengh,  and 
the  connections  should  develop  the  full  strength  of  the  body  of 
the  members  connected.  The'  connections  should  be  as  direct 
as  possible.  When  a  live  load  is  joined  to  a  static  load,  the 
judgment  of  the  designer,  or  of  the  one  who  prepares  the  specifi- 
cations for  the  designer,  must  be  exercised.  A  warehouse  floor 
to  be  loaded  with  a  certain  class  of  goods  has  maximum  stresses 
from  a  static  load.  The  floor  of  a  drill-room,  ball-room,  or 
highway  bridge  receives  maximum  loading  from  a  crowd  of  people 

*  A  portion  of  these  paragraphs  is  extracted  from  a  lecture  by  Mr.  C.  C.  Schneider. 


SAFE  WORKING  STRESSES.  159 

the  possible  density  of  which  can  be  ascertained.  But  if  these 
masses  of  people  keep  moving,  and  more  particularly  if  they  keep 
step,  the  effect  of  their  weight  will  be  increased  by  the  vibrations 
resulting  therefrom.  This  action  is  generally  called  impact. 

In  the  case  of  a  building,  the  floor-joists,  receiving  the  impact 
directly,  will  be  most  affected;  the  girders  which  carry  the  joists 
will  be  less  affected,  and  the  columns  which  support  the  girders 
will  receive  a  smaller  percentage  of  the  impact,  the  proportionate 
effect  growing  less  as  the  number  of  stories  below  the  given  floor 
increases.  In  the  absence  of  trustworthy  data  from  which  to 
determine  this  impact,  it  is  left  to  the  judgment  of  the  engineer 
to  increase  the  live  load  by  a  certain  percentage,  or  to  decrease 
the  allowable  unit  stress,  for  each  case,  to  provide  for  the  effect, 
as  will  be  seen  in  the  values  given  later. 

For  economy,  make  designs  which  will  simplify  the  shop  work, 
reduce  the  cost,  and  insure  ease  of  fitting  and  erection.  Avoid 
an  excess  of  blacksmith  work  and  much  use  of  bent  pieces. 

144.  Dead  and  Live  Load. — From  what  has  been  said  it  is 
readily  seen  that  a  moving  or  live  load  has  a  much  more  serious 
effect  on  a  truss  than  a  static  or  dead  load,  and  in  the  case  of  a 
railroad  bridge  the  stresses  in  the  members  due  to  a  rapidly 
moving  train  are  much  greater  than  the  stresses  would  be  under 
such  a  train  at  rest.  In  designing  the  members  of  steel  bridges 
this  increase  of  stress  is  usually  provided  for  by  one  of  three  ways: 

Using  different  working  stresses  for  dead  and  live  load,  as  a 
unit  stress  of  18,000  Ib.  per  sq.  in.  for  dead  load  and  9,000  Ib.  for 
live.  When  this  method  is  employed  the  ratio  is  commonly  2  to  i. 

Using  a  varying  unit  stress  found  from  a  modified  form  of  the 

.     e  (       min.  stressx 

Launhardt-Weyrauch  formula,  as  7  =  0,000(1-!-  — ).     If 

\       max.  stress/ 

the  stresses  reverse  the  fraction  is  negative.  The  coefficient  J  in ' 
the  original  formula  is  dropped  arbitrarily. 

Adding  a  certain  percentage  of  the  live  load,  called  impact, 
to  the  sum  of  the  dead-  and  live-load  stresses,  and  using  a  constant 
working  stress  as  18,000  Ib.,  one  impact  formula  often  used  is 

Impact  = ~= JT-X live-load  stress.      By   ''length  "  is  meant 

the  distance  in  feet  through  which  the  load  must  pass  to  produce 
maximum  stress  in  the  member  in  question.  Such  experiments 
as  have  been  made  on  bridges  under  moving  load  indicate 'that 


i6o 


STRUCTURAL  MECHANICS. 


the  impact  is  not  as  great  as  the  formula  gives.  The  third  method 
is  sometimes  combined  with  the  first  or  second,  but  with  the  use 
of  a  different  impact  formula. 

While  these  three  ways  of  designing  are  quite  different  in  theory 
and  method,  the  resulting  sizes  of  members  are  much  the  same. 

145.  Working  Stresses  for  Timber. — The  following  table  is 
from  Bulletin  No.  12,  U.  S.  Dept.  of  Agriculture,  Div.  of  Forestry: 

SAFE  UNIT  STRESSES  AT  18%  MOISTURE. 
Structures  freely  exposed  to  the  weather. 


Compression. 

Bending. 

Shear. 

Tension. 

With 
Grain. 

Across 
Grain. 

Extreme 
iMbre. 

With 
Grain. 

I,OOO 

840 
700 
76o 
880 
650 
800 

215 
215 
I5° 
140 
170 

IJ5 
400 

1.550 

1,300 
880 
1,090 
1,320 

I,20O 

125 
IOO 

75 
200 

12,000 

9,000 
7,000 

10,000 

Short-leaf  pine  
White  pine               .  .  . 

Norway  pine         .  .  . 

Douglas  spruce  
Redwood           

White  oak  

Factor  of  safety  .... 

5 

3 

5 

4 

I 

The  values  given  were  found  by  taking  the  average  of  the 
lowest  10  per  cent,  of  the  results  of  tests  on  non-defective  timber 
and  dividing  by  the  factor  of  safety  given  in  the  last  row.  As 
timber  never  fails  in  tension  the  safe  tensile  stress  is  not  given; 
the  last  column  shows  the  ultimate  strength.  For  structures 
protected  from  the  weather  the  compressive  and  bending  values 
may  be  increased  20  per  cent.  The  shearing  value  should  not 
be  increased.  The  table  was  constructed  for  designing  railway 
trestles. 

The  following  safe  working  stresses  have  been  recommended 
by  the  Committee  on  Strength  of  Bridge  and  Trestle  Timbers, 
Am.  Assn.  of  Ry.  Supts.  of  Bridges  and  Buildings : 


SAFE  WORKING  STRESSES. 


161 


Tension. 

Compression. 

Bend- 
ing. 

Shear. 

With 
Grain. 

Across 
Grain. 

With  Grain. 

Across 
Grain. 

Ex- 
treme 
.Fibre. 

With 
Grain. 

Across 
Grain. 

End 
Bear- 
ing. 

Cols, 
under 

is 

Diams. 

1,200 
700 
800 
1,200 
800 
600 
700 
1,000 

60 
5° 

50 

1,  600 

1,100 
1,200 

1,  600 

1,200 

1,000 
700 
800 

1,200 
800 
800 
800 
QOO 

35° 
200 
200 
300 
200 

150 
200 
500 

1  200 
700 
700 
1,100 
700 
600 

750 
1,000 

15° 

IOO 

150 

IOO 

IOO 
IOO 
200 

1,250 
500 

750 
600 

1,000 

Douglas  spruce 

Spruce  and  Eastern  fir 
Hemlock  
Redwood 

White  oak  

200 

1,400 

Factor  of  safety  

IO 

IO 

5 

5 

4 

6 

4 

4 

Messrs.  Kidwell  and  Moore  deduced  the  following  column 
formulas  based  on  the  values  given  in  the  preceding  table.  Least 
cross-dimension  of  column  =  &;  length  =/. 


ijooo—nl  + 
700—    8l+ 

800-    81  + 
1,200—  i2l+ 


Long-leaf  pine  ..........  1,000-8.5/^/2, 

White  pine  .............       joo—6.  l+h 

Norway  pine,  spruce,  and 

Eastern  fir  ..........       8oo-6.$l+h 

Douglas  spruce.  .  .......  1,200-9.  l+h 

Hemlock  ...............      8oo  —  >j.6l+h 

Redwood  ..............      800  -8.61  +h  800  --  nl  +h 

Oak  ...................       goo—q.l-r-h  qoo—nl+h 

For  buildings  Mr.  C.  C.  Schneider  recommends  the  following 
unit  stresses: 


Compression 

Bending. 

With 

Grain. 

Shear  with 
Grain. 

End 
Bearing. 

Columns 
under  10 
Diameters. 

Across 
Grain. 

Extreme 
Fibre. 

I,<OO 

I  OOO 

•7  £Q 

T  rno 

White  pine,  spruce.  .. 

I,OOO 

800 

600 

CQO 

200 

I,  OOO 

800 

IOO 

White  oak  

I  200 

I  OOO 

C.OO 

I  2OO 

s,UU 

1 62  STRUCTURAL  MECHANICS. 

P  f     I 

For  columns  when  l+h>io,  -zr=f 7-,  in  which  /   is 

S          100  h 

the  compressive  stress  for  a  short  column  taken  from  the  table. 

The  strength  of  columns  built  up  by  bolting  two  or  more 
pieces  of  timber  together  is  no  greater  than  that  of  the  individual 
pieces  acting  independently. 

The  names  of  woods  vary  in  different  localities;  long-leaf, 
short-leaf,  and  one  or  two  other  Southern  species  of  pine  are  not 
distinguished  in  the  market  and  are  called  Southern,  Georgia,  or 
yellow  pine.  Douglas  spruce  is  also  known  as  Oregon  fir,  spruce, 
or  pine. 

146.  Working  Stresses  for  Railroad  Bridges. — Mr.  Theodore 
Cooper's  Specifications,  edition  of  1901,  give  the  following  unit 
stresses  in  pounds  per  square  inch. 

MEDIUM    STEEL   IN   TENSION. 

Main  truss  members — dead  load 20,000 

live      "    10,000 

Floor-beams  and  stringers 10,000 

Lateral  and  sway  bracing — wind  load 18,000 

"      "          "           live      "    12,000 

Floor -beam  hangers  and  members  liable  to  sudden  loading  6,000 

For  sojt  steel  in  tension  reduce  the  unit  stresses  10  per  cent. 

MEDIUM   STEEL    IN    COMPRESSION, 

Chord  segments — dead  load 20,000  —go/  -r- r 

11  "  live  li  10,000— 45 /-hr 

Posts  of  through-bridges — dead  load 1 7,000  —gol  +r 

"  "  "  live  "  8,500 -45/+r 

Posts  of  deck-bridges  and  trestles — dead  load 18,000  —  8o/  -±r 

"  "  "  '.'  live  "....  9,ooo-4o/^r 

Lateral  and  sway  bracing — wind  load 13,000  —6ol+r 

11  li  live  "  8,7oo-4o/-hr 

For  sojt  steel  in  compression  reduce  the  unit  stresses  15  per 
cent. 

The  ratio  of  /  to  r  shall  not  exceed  100  for  main  members  and 
120  for  laterals. 

When  a  member  is  subjected  to  alternating  stresses,  each 
stress  shall  be  increased  by  eight-tenths  of  the  lesser  and  the 
member  designed  for  that  stress  which  gives  the  larger  section. 


SAFE  WORKING  STRESSES,  163 

PINS   AND   RIVETS. 

Shear 9,000 

Bearing  (thickness  X diameter) 1 5,000 

Bending  (pins) 18,000 

For  floor    rivets  increase  number  of  rivets  25%. 
"    field        "          "  "        "     "      50%. 

'     lateral    "     decrease      "        "     "      33%. 

PEDESTALS. 

Pressure  on  rollers,   pounds  per  lineal  inch,   3ooXdiam.  in 
inches. 

Pressure  of  bedplates  on  masonry,  pounds  per  square  inch, 

250. 

The  specifications  of  the  American  Bridge  Co.,  1900,  give  the 
following  working  stresses: 

The  maximum  stress  in  a  member  is  found  by  adding  the  dead 

load,  live  load,  and  impact.     Impact  = — yX  live-load   stress, 

in  which  L  is  length  in  feet  of  distance  to  be  loaded  to  produce 
maximum  stress  in  member. 

Soft  Steel.  Medium  Steel.. 

Tension 15,000  17,000 

1^,000  17,000 

Compression — 


I2 

1  + 


n,ooor2 
Shear  on  web-plates 9,000  10,000 

The  ratio  of  /  to  r  shall  not  exceed  100  for  main  compression 
members  and  120  for  laterals. 

When  a  member  is  subjected  to  alternating  stresses  the  total 
section  shall  be  equal  to  the  sum  of  the  areas  required  for  each 

stress. 

PINS   AND   RIVETS. 

Shear 1 1,000  12,000 

Bearing  (thickness  X diameter) 22,000  24,000 

Bending  (pins) 22,000  25,000 

For  hand-driven  field  rivets  increase  number  25%. 
"    power -driven   "         "  "  "         10%. 


1 64  STRUCTURAL  MECHANICS. 

PEDESTALS. 

Pressure  on  rollers,  pounds  per  lineal  inch, 


i2,oooVdiam.  in  inches. 

Pressure  of  bedplates  on  masonry,  pounds  per  square  inch, 
400. 

147.  Working  Stresses  for  Highway  Bridges. — Mr.  Cooper's 
Specifications,  1901,  give  the  following  stresses: 

MEDIUM   STEEL   IN  TENSION. 

Main  truss  members — dead  load 25,000 

live      "    12,500 

Floor-beams,  stringers,  and  riveted  girders 13,000 

Lateral  and  sway  bracing — wind  and  live  load 18,000 

Floor-beam  hangers  and  members  liable  to  sudden  loading      8,000 
For  soft  steel  in  tension  reduce  the  unit  stresses  10  per  cent. 

MEDIUM   STEEL   IN   COMPRESSION. 

Chord  segments — dead  load 24,000  —  i  lol  —r 

"  ll  live     "   12,500—  55/— r 

Posts  of  through-bridges — dead  load 20,000  —  gol—r 

"      "  "  live       ".... 10,000-  45/-r 

Posts  of  deck-bridges — dead  load 22,000  —  8o/  —  r 

live      "    11,000-  4ol+r 

Lateral  and  sway  bracing — wind  load 13,000  —  6ol+r 

live        "    8,700-  4o/H-r 

For  soft  steel  in  compression  reduce  the  unit  stresses  15  per 
cent. 

The  ratio  of  /  to  r  shall  not  exceed  100  for  main  members  and 
1 20  for  laterals. 

When  a  member  is  subjected  to  alternating  stresses,  each 
stress  shall  be  increased  by  eight-tenths  of  the  lesser  and  the 
member  designed  for  that  stress  which  gives  the  larger  section. 

PINS   AND   RIVETS. 

Shear 10,000 

Bearing  (diameter  X  thickness) 18,000 

Bending  (pins) 20,000 

For  floor     rivets  increase  number  25%. 
"    field          "  "  "       50%. 

"    lateral      <(      decrease      "       3°%- 


SAFE  WORKING  STRESSES.  165 

PEDESTALS 

Pressure  on  rollers,  pounds  per  lineal  inch,  3ooXdiam.  in 
inches. 

Pressure  of  bed-plates  on  masonry,  pounds  per  square  inch, 

250. 

TIMBER   FLOOR-JOISTS. 

Fibre  stress  on  yellow  pine  or  white  oak 1,200 

*  *      "   white  pine  or  spruce 1,000 

The  working  stresses  given  by  the  American  Bridge  Co. 
Specifications  for  Highway  Bridges,  1901,  are  the  same  as  for 
Railway  Bridges,  with  the  following  exceptions: 

The  sum  of  dead  and  live  load  is  to  be  increased  by  25  per 
cent,  of  the  live  load  to  compensate  for  the  effect  of  impact  and 
vibration. 

The  ratio  of  /  to  r  shall  not  exceed  120  for  main  members  and 
140  for  laterals. 

When  a  member  is  subjected  to  alternating  stresses,  design 
for  that  stress  which  gives  the  larger  section. 

Fibre  stresses  on  floor- joists  same  as  in  preceding  specification. 

148.  Working  Stresses  for  Buildings. — A  set  of  building  specifi- 
cations by  Mr.  C.  C.  Schneider  may  be  found  in  the  Transactions 
of  the  American  Society  of  Civil  Engineers,  Vol.  LIV,  June, 
1905.  Some  of  the  working  stresses  recommended  are  abstracted 
below. 

Structural  steel  of  ultimate  strength  between  55,000  and 
65,000  Ib.  per  sq.  in.  is  to  be  used. 

Tension 16,000 

Compression 16,000  —  jol-^-r 

Shear  on  web  plates  (gross  section) 10,000 

For  wind  bracing  and  the  combined  stresses  due  to  wind  and 
the  other  loading  the  permissible  stresses  may  be  increased  25 
per  cent. 

The  ratio  of  /  to  r  shall  not  exceed  125  for  main  compression 
members  and  150  for  wind  bracing. 

Members  subjected  to  alternating  stresses  shall  be  propor- 
tioned for  the  stress  giving  the  larger  section. 


166  STRUCTURAL  MECHANICS. 

PINS   AND   RIVETS. 

Shear  on  pins  and  rivets 12,000 

"      "  bolts 9,000 

Bearing  on  pins  and  rivets 24,000 

' '  bolts 18,000 

Bending  on  pins 24,000 

For  field  rivets  increase  number  33  per  cent. 

WALL   PLATES    AND   PEDESTALS. 

Bearing  pressures  in  pounds  per  square  inch  on 

Brickwork,  cement  mortar 200 

Rubble  masonry,  cement  mortar 200 

Portland  cement  concrete 350 

First-class  masonry,  sandstone 400 

granite 600 


<  <  it 


MASONRY. 

Permissible  pressure  in  masonry,  tons  per  square  foot : 

Common  brick,  natural  cement  mortar 10 

1 1      Portland  cement  mortar 12 

Hard-burned  brick,  Portland  cement  mortar 15 

Rubble  masonry,  Portland  cement  mortar 10 

Coursed  rubble  "     12 

Portland  cement  concrete,  1-2-5 20 

149.  Machinery,  etc. — The  designing  of  machinery  has  not 
been  systematized  as  has  the  designing  of  structural  steel  work, 
and  the  choice  of  working  stresses  is  to  a  large  extent  a  matter  of 
individual  judgment.  The  following  values  are  taken  from  a 
table  in  Un win's  Machine  Design  and  are  for  unvarying  stress; 
if  the  stress  varies,  but  does  not  reverse,  multiply  by  two-thirds; 
if  the  stress  reverses,  multiply  by  one- third. 

Tension.          Compression.         Torsion. 

Cast  iron 4,200  12,600  2,100 

Iron  forging 15,000  15,000  6,000 

Mild  steel 20,000  20,000  8,000 

Cast  steel 24,000  24,000  10,000 

For  shafting  of  marine  engines  9,000  Ib.  per  sq.  in.,  on  wrought 
iron  and  12,000  Ib.  on  steel  is  commonly  used,  the  shaft  being 
designed  for  the  maximum  and  not  the  mean  twisting  moment. 
These  values  are  about  those  used  in  other  machinery. 

The  working  stress  in  boilers  made  of  60,000  Ib.  steel  may 
be  as  high  as  13,000  Ib.  per  sq.  in.  with  first-class  workmanship, 
but  ordinarily  10,000  to  11,000  Ib.  is  employed. 


CHAPTER  XI. 
INTERNAL  STRESS:    CHANGE  OF  FORM. 

150.  Introduction. — Let  any  body  to  which  external  forces 
are  applied  be  cut  by  a  plane  of  section.  The  force  with  which 
one  part  of  the  body  acts  against  the  other  part  of  the  body  is  the 
stress  on  the  plane  of  section.  This  stress  is  distributed  over  the 
section  and  may  be  uniform  or  may  vary.  Its  intensity,  that  is, 
the  unit  stress,  at  any  point  is  found  by  dividing  the  stress  on  an 
indefinitely  small  area  surrounding  the  point  by  the  area.  Stresses 
may  be  normal,  tangential  or  oblique  to  the  surface  on  which  they 
act,  and  they  are  completely  determined  only  when  both  intensity 
and  direction  are  known.  It  is  desirable  to  know  the  magnitude 
and  kind  of  the  unit  stress  at  each  point  in  order  to  be  sure  that 
the  material  can  safely  resist  it;  or  to  determine  the  required 
cross-section  to  reduce  the  existing  stresses  to  safe  values. 

A  unit  stress  is  expressed  as  a  certain  number  of  pounds  of 
tension,  compression,  or  shear  on  a  square  inch  of  section.  If 
the  plane  of  section  is  changed  in  direction,  the  force  on  the 
section  may  be  changed  and  the  area  of  section  may  also  be  changed 
so  that  the  unit  stress  on  the  new  section  is  altered  from  that  on 
the  old  in  two  ways.  Stresses  per  square  inch,  or  unit  stresses, 
therefore  cannot  be  resolved  and  compounded  as  can  forces, 
unless  they  happen  to  act  on  the  same  plane.  Generally,  each 
unit  stress  may  be  multiplied  by  the  area  over  which  it  acts,  and 
the  several  jorces  so  obtained  may  be  compounded  or  resolved 
as  desired ;  the  final  force  or  forces  divided  by  the  areas  on  which 
they  act  will  give  the  desired  unit  stresses. 

Where  the  stress  on  a  plane  varies  from  point  to  point,  as 
does  the  direct  stress  on  the  right  section  of  the  I  earn,  and  as  does 
the  shearing  stress  also  in  the  same  case,  the  investigation  is 
supposed  to  be  confined  to  so  small  a  portion  of  the  body  that 
the  stress  over  any  plane  may  be  considered  to  be  sensibly  constant. 

167 


168 


STRUCTURAL  MECHANICS. 


Fig.  2 


151.  Stress  on  a  Section  Oblique  to  a  Given  Force. — Suppose 
a  short  column  or  bar  to  carry  a  force  of  direct  compression  or 

tension,  of  magnitude  P,  cen- 
trally applied  and  uniformly 
distributed  over  the  cross- 
section,  S.  The  unit  stress  on 
and  perpendicular  to  the  right 
section  will  be  pi^P-^S. 

On  an  oblique  section  CD, 
Fig.  2,  making  an  angle  0 
with  the  right  section  A  B, 
the  unit  stress  will  be  P  +S 
sec  0  =  pi  cos  dj  making  an 

angle  of  6  with  the  normal  to  the  oblique  section  on  which 
it  acts.  If  this  oblique  unit  stress  is  resolved  normally  and 
tangentially  to  C  D,  the 

Normal  unit  stress  =  pn  =  pi  cos  #-cos  6  =  pi  cos2  #; 
Tangential     do.      =  q  =pi  cos  0  sin  0. 

The  normal  unit  stress  on  the  oblique  plane  is  of  the  same 
kind  as  P,  tension  or  compression;  the  tangential  unit  stress,  or 
shear,  tends  to  make  one  part  of  the  prism  slide  down  and  the 
other  part  slide  up  the  plane. 

The  largest  normal  unit  stress  for  different  planes  is  found 
when  0  =  o,  which  defines  the  fracturing  plane  for  tension;  the 
minimum  normal  unit  stress  occurs  for  #  =  90°;  and  the  greatest 
unit  shear  is  found  for  #  =  45°,  when  we  have  $rTOOx.  =  i#i' 

152.  Combined  Stresses. — The  action  line  of  P  may  be  taken 
for  the  axis  of  X.     Two  equal  and  opposite  forces,  pull  or  thrust, 
may  then  be  applied  along  the  axis  of  Y,  and  the  normal  and 
tangential  unit  stresses  found  on  the  plane  just  discussed;    and 
similarly  for  the  direction  Z.     The  normal  unit  stresses,  since 
they  act  on  the  same  area,  may  then  be  added  algebraically,  and 
the  shearing  stresses  may  be  combined;    finally  a  resultant  ob- 
lique unit  stress  may  be  found  on  the  given  plane. 

Thus,  if  Pi  is  the  unit  stress  on  a  section  normal  to  the  X 
axis,  and  p2  the  unit  stress  on  a  section  normal  to  the  Y  axis, 


INTERNAL  STRESS:    CHANGE  OF  FORM.  169 

pn  =  pi  cos2  0  +  p2  cos2(9o°-0)  =  pl  cos2  0  +  p2  sin2  0, 
q  =pi  sin  #  cos  6  —  p2  cos  0  sin  0  =  (/>i  —p2)  sin  0  cos  0. 

pn  is  a  maximum  when  6  =  0  or  90°,  and  g  is  a  maximum  when 

0=45°. 


A  more  convenient  method  will,  however,  be  developed  and 
used  in  the  following  sections.  As  most  of  the  forces  which  act 
on  engineering  structures  lie  in  one  plane  or  parallel  planes,  such 
cases  chiefly  will  be  considered. 

153.  Unit  Shears  on  Planes  at  Right  Angles.  —  If,  in  the  pre- 
ceding illustration,  the  unit  stresses,  both  normal  and  tangential, 
are  found  on  another  plane  N  which  makes  an  angle  6'  =  90°—  6 
with  the  right  section,  there  will  result 


Hence,  on  a  pair  of  planes  of  section,  both  of  which  are  at  right 
angles  to  the  plane  of  external  forces  and  to  each  other,  the  unit 
shears  are  0}  equal  magnitude,  and,  since  pn  +  pn=pi,  the  unit 
normal  stresses  are  together  equal  to  the  original  normal  unit 
stress.  It  is  further  evident  that  one  normal  unit  stress  pnf  may 
be  found  by  subtraction  as  soon  as  the  other  is  known,  and  that 
ordinary  resolution  on  these  two  planes  of  the  original  unit  stress 
would  be  erroneous. 

154.  Unit  Shears  on  Planes  at  Right  Angles  Always  Equal.  — 
Since,  as  before  stated,  other  forces,  in  other  directions,  may  be 
simultaneously  applied  to  the  given  A 

body,  and  their  effects  found  on  the  ' 

same  two  planes,  it  follows  that,  in 
any  body  under  stress,  equal  unit 
shearing  stresses  will  exist  on  two 
planes  each  of  which  is  perpendicular 
to  the  direction  of  the  shear  on  the 
other.  I 

Example.  —  A   closed   cylindrical   re-  Fig.  61 

ceiver,  J  in.  thick,  has  a  spiral  riveted 
joint  making  an  angle  of  30°  with  the  axis  of  the  cylinder,  and 


1  70  STRUCTURAL  MECHANICS. 

a  portion  2  in.X4  in.  of  the  cylinder,  Fig.  61,  has  the  given  tensions  of 
2,500  Ib.  acting  upon  it.  Then  ^1  =  2,500-^2-  £=5,000  Ib.  per  sq.  in., 
and  />2=  2,  500  -7-4-  £=2,500  Ib.  per  sq.  in. 

/>n=5,ooo-f  +2,500-  £=4,375  Ib.  per  sq.  in., 
q=  5,000-0.433-2,500-0.433  =  1,082  Ib.  per  sq.  in., 

.  per  sq.  in., 


or  4,507-  £=1,127  Ib.  per  linear  inch  of  joint,  which  value  will  deter- 
mine the  necessary  pitch  of  the  rivets  for  strength. 

The  stress  on  a  joint  at  right  angles  to  the  above  can  be  similarly 
found.  An  easier  process  will  be  given  in  §  161. 

JSS-  Compound  Stress  is  the  internal  state  of  stress  in  a  body 
caused  by  the  combined  action  of  two  or  more  simple  stresses 
(or  balanced  sets  of  external  forces)  in  different  directions,  as  in 
the  above  example.  The  investigations  which  follow  are  those 
of  compound  stress,  but  they  will,  as  above  stated,  be  chiefly 
confined  to  stresses  in  or  parallel  to  one  plane. 

156.  Conjugate  Stresses  :  Principal  Stresses.  —  If  the  stress  on 
a  given  plane  in  a  body  is  in  a  given  direction,  the  stress  on  a 
plane  parallel  to  that  direction  must  be  parallel 
to  the  first-mentioned  plane.  For  the  equal  re- 
sultant forces  exerted  by  the  other  parts  of  the 
body  on  the  faces  A  B  and  C  D  of  the  pris- 
matic particle,  Fig.  62,  are  directly  opposed  to 
one  another,  their  common  line  of  action  trav- 
ersing the  axis  of  X  through  O;  and  they  are 
therefore  independently  balanced.  Therefore 
the  forces  exerted  by  the  other  parts  of  the  body  on  the  faces  A  D 
and  B  C  of  this  prism  must  be  independently  balanced  and 
have  their  resulcants  directly  opposed;  which  cannot  be  unless 
their  direction  is  parallel  to  the  plane  Y  O  Y. 

A  pair  of  stresses,  each  acting  on  a  plane  parallel  to  the  direc- 
tion of  the  other,  is  said  to  be  conjugate.  Their  unit  values  are 
independent  of  each  other,  and  they  may  be  of  the  same  or  oppo- 
site kinds.  If  they  are  normal  to  their  planes,  and  hence  at 
right  angles  to  each  other,  they  are  called  principal  stresses. 

Examples.  —  The  unit  stress  found  in  §  154  makes  an  angle  with  the 
plane  on  which  it  acts  whose  tangent  is  4,375-5-1,082  =  4.04.  Upon  a 


INTERNAL  STRESS:    CHANGE  OF  FORM. 


171 


new  plane  cutting  the  metal  in  this  direction  the  stress  must  act  in  a 
direction  parallel  to  the  joint  referred  to. 

If  a  plane  be  conceived  parallel  to  a  side-hill  surface,  at  a  given 
vertical  distance  below  the  same,  the  pressure  at  all  points  of  that 
plane,  being  due  to  the  weight  of  the  prism  of  earth  above  any  square 
foot  of  the  plane,  will  be  vertical  and  uniform.  Then  must  the  pressure 
on  a  vertical  plane  transverse  to  the  slope  be  parallel  to  the  surface  of 
the  ground.  That  the  pressure  against  the  vertical  plane  is  not  hori- 
zontal, but  inclined  in  the  direction  stated,  is  shown  by  the  movement 
of  sewer  trench  sheeting  and  braces,  when  the  braces  are  not  inclined 
up  hill,  but  are  put  in  horizontally. 

157.  Shearing  Stress.  —  If  the  stresses  on  a  pair  of  planes  are 
entirely  tangential  to  those  planes,  the  unit  shears  must  be  equal. 
Consider  them  as  acting  along  the 
faces  of  a  small  prismatic  particle 
A  BCD,  which  lies  at  O.  The 
moment  of  the  total  shear  on  the 
two  faces  A  B  and  C  D  must  bal- 
ance the  moment  for  the  faces  A  D 
and  B  C,  for  equilibrium.  Fig.  63 


But  the  area  of  A  B  C  D,  AB-EF=AD-HG;     /.  f  =  q. 

This  construction  shows  further  that  a  shear  cannot  act  alone 
as  a  simple  stress,  but  must  be  combined  with  an  equal  unit  shear 
on  a  different  plane. 

158.  Two  Equal  and  Like  Principal  Stresses.  —  If  a  pair  of 
principal  stresses,  §  156,  are  equal  unit  stresses  of  the  same  kind, 
pi  and  p2,  Fig.  64,  the  stress  on  every  plane  is  normal  to  that 
plane,  and  of  the  same  kind  and  magnitude. 


-p.      -P, 


K 


°v 


Fig.  64 


Let  Pi  act  in  the  direction  O  X  on  the  plane  O'  B'  of  the 
prismatic  particle  O'  A'  B'  which  lies  at  O,  and  p2  act  in  the  direc- 


172  STRUCTURAL   MECHANICS. 

tion  O  Y  on  the  plane  O'  A',  pi  and  p2  being  equal  unit  stresses 
of  the  same  kind.  Make  OD=^i-O'B',  the  total  force  on 
O'B',  and  O  E  =  p2-O'  A',  the  total  force  on  O'  A',  both  being 
positive.  Complete  the  rectangle  O  D  R  E.  Then  must  R  O, 
applied  to  the  plane  A'B',  be  necessary  to  insure  equilibrium 
of  the  prism  O'  A'  B'.  Hence  /  =  R  O  n-A'  B'.  Since  pi  =  p2, 

OP      OE      OR 


O'B'~O'A'~A'B'          •'      --' 

Because  of  similarity  of  triangles  A'  O'  B'  and  O  E  R,  R  O  is 
perpendicular  to  A  B,  or  ft  to  A'  B',  and  is  of  the  same  kind  as 
P\  and  p2* 

Example.  —  Fluid  pressure  is  normal  to  every  plane  passing  through 
a  given  point,  and  equal  to  the  pressure  per  square  inch  on  the  horizontal 
plane  traversing  the  point.  Here  manifestly  the  three  coordinate  axes 
of  X,  Y,  and  Z  might  be  taken  in  any  position,  as  all  stresses  are  princi- 
pal stresses. 

159.  Two  Equal  and  Unlike  Principal  Stresses.  —  If  a  pair  of 
principal  stresses  are  equal  unit  stresses  of  opposite  kinds,  as  pi 
and  -  p2,  Fig.  64,  the  unit  stress  on  every  plane  will  be  the  same 
in  magnitude,  but  the  angle  which  its  direction  makes  with  the 
normal  to  its  plane  will  be  bisected  by  the  axis  of  principal  stress, 
and  its  kind  will  agree  with  that  of  the  principal  stress  to  which 
it  lies  the  nearer. 

In  this  case  lay  off  Oe  in  the  negative  direction,  to  represent 
—p2-O"  A";  construct  the  rectangle  O  Dre,  and  draw  rO  which 
will  be  the  required  force  distributed  over  A"  B"  to  balance  the 
forces  O"  B"  and  O"  A".  This  force  rO  will  be  the  same  in 
magnitude  as  R  O,  making  pf  =  pi=p2  and  rO  will  make  the 
same  angle  with  O  X  or  O  Y  as  R  O  does.  As  R  O  lies  on  the 
normal  to  A  B,  and  O  X  bisects  R  Or,  the  statement  as  to  position 
is  proved.  The  stress  p  agrees  in  kind  with  that  one  of  the 
principal  stresses  to  which  its  direction  is  nearer. 

1  60.  Two  Shears  at  Right  Angles  Equivalent  to  an  Equal  Pull 
and  Thrust.—  If  the  plane  A"  R"  is  at  an  angle  of  45°  with  O  X, 
rO  will  coincide  with  A  B  and  becomes  a  shear.  Therefore  two 


INTERNAL  STRESS:    CHANGE  OF  FORM.  173 

equal  unit  stresses  of  opposite  kinds,  that  is  a  pull  and  a  thrust 
and  normal  to  planes  at  right  angles  to  one  another,  are  equivalent 
to,  and  give  rise  to  equal  unit  shears  on  planes  making  45°  with 
the  first  planes  and  hence  at 
right  angles  to  each  other;  and 
vice  versa. 


Fig.  65 


Example. — If,  at  a  point  in  the 
web  of  a  plate  girder,  Fig.  65,  there 
is  a  unit  shear,  and  nothing  but 
shear,  on  a  vertical  plane,  of  4,000 
lb.,  there  must  be  a  unit  shear  of 
4,000  lb.,  and  nothing  but  shear,  on  the  horizontal  plane  at  that  point 
and  on  the  two  planes  inclined  at  45°  to  the  vertical  through  the  same 
point  there  will  be,  on  one,  only  a  normal  unit  tension  of  4,000  lb.,  and 
on  the  other  an  equal  normal  unit  thrust. 

161.  Stress  on  any  Plane,  when  the  Principal  Stresses  are 
Given. — This  general  problem  is  solved  by  dividing  it  into  two 
special  cases ;  the  one  that  of  two  equal  and  like  principal  stresses, 
the  other  that  of  two  equal  and  unlike  principal  stresses.  Let 
the  two  principal  unit  stresses  be  pi=O  D,  and  p2  =  O  F,  of  any 
magnitude,  and  of  the  same  kind  or  opposite  kinds.  Fig.  66. 
The  direction,  magnitude  and  kind  of  the  unit  stress  on  any  plane 
A  B  is  desired. 

Let  pi  be  the  greater.  Divide  pi  and  p2  into  their  half  sum 
and  difference  as  follows : 

Pi=%(pi+p2)+%(pi-p2),     and 

The  distance  O  C  or  O  E  will  represent  the  half  sum 
%(pi+p2),  and  CD  or  EF  the  half  difference  \(p\.-p^.  If 
pi  and  p2  are  the  same  sign  the  right-hand  figure  will  result; 
if  of  opposite  signs,  the  left-hand  figure  will  be  obtained. 

By  the  principle  of  §  158,  when  the  two  equal  principal  unit 
stresses  O  C  and  O  E  are  considered,  lay  off  O  M  on  the  normal 
to  the  plane  whose  trace  is  A  B,  for  the  direction  and  magnitude 
of  the  unit  stress  on  A  B  due  to  %(P\+p2)>  There  remain 
C  D  and  E  F  representing  +%(p\—  p2)  on  the  vertical  axis, 
and  —%(pi—p2)  on  the  horizontal  axis  respectively. 


174 


STRUCTURAL  MECHANICS. 


By  §  159,  lay  off  O  Q,  making  the  same  angle  with  O  X  as 
does  O  M,  but  on  the  opposite  side,  or  in  the  contrary  direction, 
for  the  magnitude  and  direction  of  stress  on  plane  A  B  due  to 
±i(#i""#2).  As  O  M  and  O  Q  both  act  on  the  same  unit  of 
area  of  A  B,  R  O,  in  the  opposite  direction  to  their  resultant 
O  R,  will  give  the  direction  and  magnitude  of  the  unit  stress  on 
A'  B'  to  balance  pi  on  O'  B'  and  p2  on  O'  A'.  In  the  figure 
on  the  right  R  O  is  positive,  or  compression.  If,  in  the  figure 
on  the  left,  where  pi  is  +  and  p2  is  — ,  R  O  falls  so  far  to  the  left 
as  to  come  on  the  other  side  of  A  B,  it  will  agree  with  p2,  and 


Fig.  66 

be  negative  or  tension.  If  A  B  is  taken  much  more  steeply 
inclined,  such  will  be  the  case.  The  small  prisms  illustrate  the 
constructions.  If  R  O  falls  on  A  B,  it  will  be  shear.  Some 
constructions  for  different  inclinations  of  plane  A  B  will  be  help- 
ful to  an  understanding  of  the  matter. 

A  much  abbreviated  construction  is  as  follows: — Strike  a 
semicircle  from  M  on  the  normal,  with  a  radius  M  Q  =  %(pi+p2), 
and  draw  M  R  through  the  points  where  the  semicircle  cuts  the 
axes  of  pi  and  p2.  The  angle  N  M  R  is  thus  double  the  angle 
MOD,  since  it  is  an  exterior  angle  at  the  vertex  of  an  isosceles 
triangle.  Lay  off  M  ^  =  ^(pi—p2)  in  the  direction  of  the  axis  of 
greatest  stress,  and  R  O  will  be  the  desired  unit  stress  on  A  B. 
If  p2  is  opposite  in  kind  to  pi,  M  R  will  be  greater  than  M  O, 
and  R  will  go  beyond  P. 


INTERNAL  STRESS:    CHANGE  OF  FORM.  175 

162.  Ellipse  of  Stress.  —  For  different  planes  A  B  through 
O,  pi  and  p2  being  given,  the  locus  of  M  is  a  circle  of  radius 
%(pi+p2)>  and  the  locus  of  R  is  an  ellipse  (as  will  be  proved 
below),  with  major  and  minor  semidiameters  pi  and  p%.  Hence 
it  is  seen  why  pi  and  p2,  normal  to  the  respective  planes,  and  at 
right  angles,  are  called  principal  stresses. 

If  three  principal  stresses,  coinciding  in  direction  with  the 
rectangular  axes  of  X,  Y  and  Z,  simultaneously  act  on  a  given 
point,  an  ellipsoid  constructed  on  them  as  semidiameters  will 
limit  and  determine  all  possible  stresses  on  the  various  planes 
which  can  be  passed  through  that  point  in  the  body. 

That  the  locus  of  R  is  an  ellipse  may  be  proved  as  follows: 
Drop  a  perpendicular  R  S  from  R  on  O  X.  P  M  O  and  O  M  G 
are  isosceles  triangles.  <POM=<MPO=0. 


R  S  =P  R  sin  MP  O=p2  sin  0, 
S  O  =  G  R  cos  MP  O=^i  cos  fl, 
O  R  =  pr  =V(S  O2  +  R  S2)  =\  /(pj2  cos2  d  +  p22  sin2  0),     .     (i) 


which  is  the  value  of  the  radius  vector  of  an  ellipse  in  terms  of 
the  eccentric  angle,  the  origin  being  at  the  centre. 

As  <NMR  =  2-<POM  =  20, 

smMOR:sin  O  M  R  =  R  M:O  R  =  J(#i  -#2):  fa 

.'.  sinMOR  =  sin  26-^—-^,    .....     (2) 

2pr 

which  gives  the  obliquity  of  the  unit  stress  to  the  normal  to  the 
plane,  in  terms  of  the  angle  of  the  normal  with  the  axis  of  greater 
principal  stress,  or  of  the  plane  with  the  other  axis.  The  graphical 
construction  gives  the  stress  and  its  angle  with  the  normal  or 
the  plane  by  direct  measurement,  and  is  far  more  convenient 
and  less  liable  to  error. 

If  pi  =p2,  the  case  reduces  to  that  of  §  158  or  §  159. 

If  the  ellipse  whose  principal  semidiameters  are  pi  and  p^ 
is  given,  the  unit  stress  on  any  plane  may  briefly  be  found  by 
drawing  the  normal  to  the  plane,  laying  off  O  M  =  %(pi+p2)> 


176  STRUCTURAL  MECHANICS. 

taking  a  radius  of  %(pi  -p2),  and,  with  M  as  a  centre,  cutting 
the  ellipse  at  R  on  the  side  of  the  normal  towards  the  greater 
axis.  A  line  R  O  will  be  the  desired  unit  stress. 

Example. — Let  ^  =  100  Ib.  on  sq.  in.,  p2=—  50  Ib.  Plane  AB 
makes  30°  with  direction  of  p2,  or  its  normal  makes  30°  with  pi.  Con- 
struct the  figure  and  find  the  magnitude,  direction,  and  sign  of  the  unit 
stress  on  the  oblique  plane.  Try  other  values. 

163.  Shearing  Planes. — To  determine  the  angle  of  the  planes 
on  which  there  is  only  shear,  and  the  conditions  which  render 
shearing  planes  possible. 

If  the  plane  A  B  of  the  previous  figure  is  to  be  the  shearing 
plane,  there  must  be  no  normal  component  upon  it,  and  therefore 
from  §  151,  if  the  plane  makes  an  angle  Os  with  p2,  or  the  normal 
to  it  is  inclined  at  an  angle  68  to  pi, 


pn  =  pi  cos2  0,  +  p2  sin2  08  =o. 
sin  08 


*  *  cos  6S 

No  shearing  plane  is  possible  unless  pi  and  p2  differ  in  sign. 
There  will  then  be  two  planes  of  shear  making  equal  angles  with 
the  direction  of  p2  or  of  pi. 

In    the    above    example,    \/(ioo  -j- 50)  =\/2=tan  0S  =  1.4142. 

0  =  54°  44'- 

If  the  ellipse  of  stress  is  drawn,  take  a  radius  equal  to  the 
side  of  a  right  angled  triangle  whose  other  side  is  %(pi  +  p%) ,  and 
hypothenuse  is  ^(pi—p2)>  and  strike  a  circle  from  the  centre  of 
the  ellipse.  Planes  drawn  through  the  points  of  intersection  of 
this  circle  and  ellipse  and  the  centre  will  be  the  shearing  planes. 
Unless  pi  and  p2  differ  in  sign,  the  circle  will  be  imaginary.  The 
value  of  the  shear  on  these  planes  is 


164.  Given  any  Two  Stresses:   to  Find  Principal  Stresses. — 

As,  in  actual  practice,  two  oblique  unit  stresses  on  different  planes 
may  often  be  known  in  magnitude,  direction  and  sign,  it  will  be 
required  to  find  the  principal  unit  stresses,  since  one  of.  them  is 


INTERNAL  STRESS:    CHANGE  OF  FORM. 


177 


the  maximum  stress  to  be  found  on  any  plane,  and  the  other  is 
the  minimum  stress  of  the  same  kind,  or  the  maximum  normal 
stress  of  the  opposite  kind. 

Given  two  existing  unit  stresses,  p  and  p'  of  any  direction,  mag- 
nitude and  sign,  to  determine  the  principal  unit  stresses,  pi  and  p2- 

If  pi  and  p2  were  known,  and  p  and  $  were  then  to  be  found 
from  the  former,  the  construction  shown  in  Fig.  67  would  be 
made,  in  which  O  M  =  O  M.'  =  %(pi+p2)  and  MR  =  M'R' 
=  ^(pi—p2)>  If  one  of  these  normals  were  revolved  around  O 
to  coincide  with  the  other,  the  point  M'  would  fall  on  M,  but 
M'  R'  would  diverge  from  M  R,  while  equal  in  length  to  it. 

Hence,  when  p  and  pf  are  the  given  quantities,  let  A  B,  Fig. 
68,  represent  the  trace  of  the  plane  on  which  p  acts,  O  N  the 


Fig.  67 


Fig.  68 


normal  to  that  plane,  and  O  R  the  unit  stress  p  in  magnitude 
and  actual  direction  of  action  on  A  B.  OR  represents  either 
tension  or  compression,  as  the  case  may  be.  Now  let  the  plane 
on  which  p'  acts,  together  with  its  normal  and  pr  itself  in  its  relative 
position,  be  revolved  about  O  until  it  coincides  with  A  B.  Its 
normal  will  fall  on  O  N  and  pr  will  be  found  at  O  R',  on  one 
side  or  the  other  of  O  N,  if  it  is  of  the  same  kind  with  p\  or  it 
is  to  be  laid  off  on  the  dotted  line  below,  if  of  the  opposite  sign. 
In  other  words,  lay  off  j/  from  O,  at  the  same  angle  with  O  N 
which  it  makes  with  the  normal  to  its  own  plane.  It  is  well,  for 
accuracy  of  construction,  to  draw  it  on  the  same  side  of  the 
normal  as  p,  the  result  being  the  same  as  if  it  were  drawn  on 
the  other  side.  (The  change  from  one  side  of  the  normal  to 
the  other  simply  consists  in  using  the  corresponding  line  on  the 


178  STRUCTURAL  MECHANICS. 

other  side  of  the  main  axis  of  the  ellipse  of  stress).  Thus  is 
found  O  R'  or  —OR'  as  the  case  may  be.  Draw  R  R'  and,  since 
R  M  R'  must  be  an  isosceles  triangle,  bisect  R  R'  at  T  and  drop 
a  perpendicular  to  R  R'  from  T  on  to  the  normal,  cutting  the 
latter  at  M.  Then  since,  as  previously  pointed  out,  O  M  = 
?(pi+p2)  and  MR  =  MR'  =  J(^i-^2),— with  M  as  a  centre, 
and  radius  M  R,  describe  a  semicircle ;  O  N  will  be  pi  and  O  S 
will  be  p2>  Since  p  is  in  its  true  position,  and  the  angle  N  M  R  = 
2  M  O  D  of  Fig.  66  or  2  M  O  X  of  Fig.  67,  the  direction  of  the 
axis  X  along  which  pi  acts  will  bisect  N  M  R,  and  the  axis  along 
which  p2  acts  will  be  perpendicular  to  axis  X.  They  may  be  now 
at  once  drawn  through  O,  if  desired. 

165.  From  any  Two  Stresses  to  Find  Other  Stresses. — From 
the  preceding  construction,  §  164,  the  stress  on  any  other  plane 
may  now  be  found.     All  possible  values  of  p  consistent  with  the 
two,  O  R  and  O  R',  first  given,  will  terminate,  in  Fig.  68,  on  the 
semicircle  just  drawn,  as  at  R",  and  the  greatest  possible  ob- 
liquity to  the  normal  to  any  plane  through  O  will  be  found  by 
drawing  from  O  a  tangent  to  this  semicircle. 

1 66.  When  Shearing  Planes  are  Possible. — In  case  the  lower 
end  of  the  semicircle  cuts  below  O,  Fig.  69,  pi  and  p2  are  of 
opposite  signs,  all  obliquities  of  stress  are  possible,  and  the  dis- 
tance from  O  to  the  point  where  the  semicircle  cuts  A  B,  being 

perpendicular  to  the  normal  O  N,  gives  the  unit 
shear  on  the  shearing  planes.  If  pi  and  p2  are 
drawn  through  O  in  position,  and  the  ellipse  of 
stress  is  then  constructed  on  them  as  semi- 
diameters  (as  can  be  readily  done  by  drawing 
two  concentric  circles  with  pi  and  p2  respectively 
as  radii,  and  projecting  at  right  angles,  parallel 
to  pi  and  p2,  to  an  intersection,  the  two  points  where  any  radius 
cuts  both  circles),  an  arc  described  from  O,  with  a  radius  equal 
to  this  unit  shear  and  cutting  the  ellipse,  will  locate  a  point  in 
the  shearing  plane  which  may  then  be  drawn  through  that  point 
and  O.  Two  shearing  planes  are  thus  given,  as  was  proved  to 
be  necessary,  §  157. 

The  above  solution  may  be  considered  the  general  case. 


INTERNAL  STRESS:   CHANGE  OF  FORM. 


179 


167.  From   Conjugate  Stresses  to  Find  Principal  Stresses. — 
If  p  and  $  are  conjugate  stresses,  it  is  evident,  from  definition, 
and  from  Fig.  62,  that  they  are  equally  inclined  to  their  respective 
normals.     Hence  O  R'  will  fall  on  O  R,  when  revolved,  both  O  R 
and  O  R'  lying  above  O  when  of  the  same  sign,  and  on  opposite 
sides  of  A  B  when  of  opposite  signs.     The  rest  of  the  construction 
follows  as  before,  being  somewhat  simplified. 

It  may  be  noted  that,  when  p\  =  ±  p2,  the  propositions  of 
§§158  and  159  are  again  illustrated. 

One  who  is  interested  in  a  mathematical  discussion  of  this 
subject  is  referred  to  Rankine's  " Applied  Mechanics,"  where  it 
is  treated  at  considerable  length.  This  graphical  discussion  is 
much  simpler,  less  liable  to  error,  and  determines  the  stresses  in 
their  true  places. 

1 68.  Stresses  in  a  Beam  — The  varying  tensile  and  compres- 
sive  stresses  on  any  section  of  a  beam  are  accompanied  by  vary- 
ing shearing  stresses  on  that  section  and  by  equal  shears  on  a 
longitudinal  section.     The  direct  or  normal  stresses  due  to  bend- 
ing moment  vary  uniformly  from  the  neutral  axis  either  way, 
§  62 ;   the  shears  are  most  intense  at  the  neutral  axis,  §  72.     The 
normal  and  shearing  stresses  on  the  cross-section  of  a  rectangular 
beam  and  also  the  resultant  stresses  on  the  section  are  shown  in 


I 


Fig.  70 


Fig.  70,  a  and  b.  The  maximum  and  minimum  unit  stresses 
(that  is,  the  principal  stresses)  at  any  point,  with  their  directions, 
may  be  found  by  a  slight  modification  of  §  164  as  follows : 

Let  Fig.  71  represent  the  stresses  acting  on  a  particle,  as  the 


i8o 


STRUCTURAL  MECHANICS. 


one  at  O,  Fig.  70.     Lay  off  O  N=pn  on  the  normal  to  the  plane 
A  B,  on  which  it  acts  in  its  true  position.     As  q  acts  on  the  same 


Fig.  71 

area  of  A  B  as  does  pn,  lay  off  R  N  =  <?  at  right  angles.  The  two 
component  stresses  being  R  N  and  N  O,  a  line  from  R  to  O  will 
represent  the  direction  and  magnitude  of  the  unit  stress  on  A  B. 
Revolve  the  horizontal  plane  through  90°  to  coincide  with  A  B. 
Then  lay  off  R'  O  =  q  as  shown.  (As  explained  in  §  164,  R'  O  may 
be  laid  off  on  the  other  side  of  the  normal,  O  N,  but  that  construc- 
tion is  not  quite  so  simple.)  Since  R  O  and  R'  O  represent  p  and 
pr  of  the  preceding  sections,  connect  R  and  R'  and  bisect  R  R'  at 
M,  which  point  falls  on  O  N  and  is  also  the  point  where  the  per- 
pendicular from  R  R'  will  strike  O  N.  Hence  pi=O  M  +  M  R 
and  p2  =  O  M  —  M  R  and  the  direction  of  pi  is  M  X,  which  bisects 
<  R  M  N.  The  principal  stresses  are  shown  in  the  figure. 

If  the  principal  stresses  are  found  at  various  points  on  the 
cross-section  of  a  loaded  beam,  the  results  will  be  as  shown  in 
Fig.  70,  c.  At  the  top  and  bottom  simple  stresses  of  compres- 
sion and  tension  exist,  while  at  the  neutral  axis  there  is  a  pull  and 
thrust  at  45°  to  the  axis,  each  equal  to  the  unit  shear  on  the  ver- 
tical or  horizontal  plane.  A  network  of  lines  intersecting  at  right 

angles,  such  as  is  shown  in  Fig.  72, 
will  give  the  direction  of  the  princi- 
pal stresses  at  any  point  of  the  beam 
and  may  be  called  lines  of  principal 
stress.  Those  convex  upward  give  the  direction  of  the  compres- 
sive  forces.  At  the  section  of  maximum  bending  moment,  at 
which  section  the  shear  is  zero,  the  curves  are  horizontal  and  the 
stress  is  greater  than  at  any  other  point  on  the  curve;  from  there 
the  stress  diminishes  to  zero  at  the  edge  of  the  beam.  The  lines 
of  principal  stress  also  show  the  traces  of  surfaces  on  which  the 
stress  is  always  normal. 


Fig.  72 


OF    THE 

UNIVERSITY 

OF 


INTERNAL  STRESS:    CHANGE  OF  FORM. 


181 


169.  Rankine's  Theory  of  Earth  Pressure.  —  Any  embank- 
ment or  cutting  keeps  its  figure  both  by  the  friction  and  by  the 
adhesion  between  the  grains  of  earth.  The  latter  is  variable  and 
uncertain  in  amount  and  may  even  be  entirely  absent,  for  which 
reason  it  is  neglected  in  the  present  discussion.  A  bank  of  earth 
is,  therefore,  to  be  conceived  as  a  granular  mass,  retaining  its 
shape  by  friction  alone,  as  is  the  case  with  a  pile  of  dry  sand. 
When  any  such  granular  material  is  heaped  up  in  a  pile,  the  sides 
slide  until  they  assume  a  certain  definite  slope  depending  upon 
the  material.  The  greatest  angle  with  the  horizontal  which  the 
sides  can  be  made  to  take  is  the  angle  of  repose,  and  its  tangent 
is  the  coefficient  of  friction  of  the  material. 

It  is  evident  that  in  a  granular  mass  there  is  pressure  on  any 
plane  which  can  be  passed  through  it,  and  from  §  161  it  is  seen 
that,  if  the  principal  stresses  in  the  mass  are  unequal,  the  stress 
on  any  plane,  not  a  principal  plane,  is  oblique.  But  in  earth,  the 
obliquity  of  the  pressure  on  any  plane  cannot  exceed  the  angle 
of  repose,  <£,  or  sliding  would  take  place  along  that  plane.  If 


Fig.  73 


Fig.  73  shows  the  plane  upon  which  the  direction  of  pressure  is 
most  oblique  (see  §  165),  it  is  necessary  for  equilibrium  that  the 
angle  ROM  shall  not  exceed  <£.  Then 


By  rearranging  the  terms,  the  ratio  between  the  principal  pres- 
sures is  found  to  be 

p2  >  i  -sin  (ft 

pi  ~~  i-f  sin  <£" 


182  STRUCTURAL  MECHANICS. 

In  the  figure,  pi  is  taken  larger  than  p2',  if  p2  is  larger  than  pi, 
the  triangle  will  lie  on  the  other  side  of  the  normal  and  the  ratio 
becomes 

p2  .  i+sin  (> 


—sn 


If  pi  is  the  principal  pressure  acting  vertically  and  due  to  the 
weight  of  the  earth  itself,  its  conjugate  pressure,  p2,  must  lie 
between  the  values  given  by  the  two  preceding  equations.  Its 
exact  amount  can  be  determined  by  the  principle  of  least  resist- 
ance, for  pi,  which  is  caused  by  the  action  of  gravity,  produces 
a  tendency  for  the  earth  to  spread  laterally,  which  tendency 
gives  rise  to  the  pressure,  p2-  As  p2  is  caused  by  pi,  p2  will 
not  increase  beyond  the  least  amount  sufficient  to  balance  pi, 
hence 

2     i  —sin  < 


pi     i+s'mfi 

and  in  Fig.  73  the  angle  R  O  M  is  equal  to  <j>. 

In  the  preceding  discussion  pi  has  been  considered  to  act  on 
a  horizontal,  and  p2  on  a  vertical  plane,  and  such  will  be  the  case 
when  the  surface  of  the  ground  is  level,  but  if  the  ground  slopes, 
the  pressure  on  a  plane  parallel  to  the  surface  is  vertical  and  the 
direction  of  pi  inclines  slightly  from  the  vertical,  as  shown  in  the 
following  example: 

Example.  —  Find  the  pressure  on  the  back  of  the  retaining-  wall 
shown  in  Fig.  74,  and  also  the  resultant  pressure  on  the  joint  C  O'.  The 
pressure  on  the  plane  O  O'  passed  parallel  to  the  surface  of  the  ground 
is  vertical  and  due  to  the  weight  of  the  earth  upon  it  of  depth  K  O. 
But  the  prism  of  earth  resting  on  one  square  foot  of  the  plane  has  a 
smaller  horizontal  section  than  one  square  foot,  and  the  ratio  of  the 
unit  vertical  pressure  on  the  plane  through  O  to  the  weight  of  a  verti- 
cal column  of  earth  one  square  foot  in  cross-section  will  be  that  of  the 
normal  O  H  to  the  vertical  O  K.  Hence  O  R(=  O  H)  represents  in  feet 
of  earth  the  pressure  per  square  foot  on  the  plane  O  O'.  Draw  a  line, 
O  S,  making  the  angle  of  repose,  <£,  with  the  normal  and  by  trial  find 
on  O  H  a  centre,  M,  from  which  may  be  drawn  a  semicircle  tangent 
to  the  line,  O  S,  and  passing  through  R.  Then  O  M=$(pi  +  p2)  and 


INTERNAL  STRESS:   CHANGE  OF  FORM. 


183 


M  R=?(pi  —  p2),  and  M  X  gives  the  direction  of  pi,  the  unit  of  pressure 
being  the  weight  of  one  cubic  foot  of  earth. 

The  principal  stresses  are  now  known  and  the  pressure  on  the  back 
of  the  wall  can  be  found.  To  find  the  pressure  on  O'  B  at  O',  draw 
O'  A  parallel  to  M  X  and  O'  M'  normal  to  the  back  of  the  wall.  Lay 
off  O'  M'  equal  to  O  M,  and  M'  R'  equal  to  M  R  and  making  the  same 
angle  with  O'  A  as  O'  M'  does.  Then  R7  O'  gives  the  direction  of  pres- 
sure on  the  back  of  the  wall  and,  when  measured  by  the  scale  of  the 


>/ 

A  i 

\ 

p/l 

w  1 

1 
1 

W         Fig 

1 
1 

M 


drawing  and  multiplied  by  the  weight  of  a  cubic  foot  of  earth,  gives 
the  pressure  in  pounds  per  square  foot  at  O'. 

As  the  pressure  on  the  back  increases  regularly  with  the  distance 
below  the  surface  of  the  ground,  the  centre  of  pressure  will  be  at  D, 
one-third  of  the  slant  height  from  O',  and  the  total  earth  pressure 
against  one  foot  in  length  of  the  wall  will  be  P=  JX  O'  B  X  O'  R'X  weight 
of  a  cubic  foot  of  earth.  If  W  is  the  weight  of  one  foot  length  of  wall 
applied  at  the  centre  of  gravity,  G,  the  resultant  of  P  and  W  is  the 
resultant  pressure  on  the  bed-joint,  O'  C,  and  the  point  where  it  cuts 
the  joint  should  lie  within  the  middle  third. 

170.  Change  of  Volume. — If  p=unit  stress  per  square  inch 
on  the  cross-section  of  a  prism,  and  A  is  the  resulting  stretch  or 


184  STRUCTURAL  MECHANICS. 

shortening  per  unit  of  length,  then  by  definition  E  =  p+A,  if 
p  does  not  exceed  the  elastic  limit. 

When  a  prism  is  extended  or  compressed  by  a  simple  longi- 
tudinal stress,  it  contracts  or  expands  laterally,  Fig.  75.  This 
contraction  or  expansion  per  unit  of  breadth  may  be  written 
TA-^w  in  which  m,  the  ratio  of  longitudinal  extension  to  lateral 
contraction,  is  a  constant  for  a  given  material,  and  for  most 
solids  lies  between  2  and  4. 

A  simple  longitudinal  tension  p  then  accompanies  a 

Longitudinal   stretch  =  X  =  p+E  per   unit    of   length,    and    a 
Transverse     contraction  =  —  X-^-  m  —  —  p-^-mE     per     unit     of 
breadth. 

For  ordinary  solids  A  is  so  small  that  it  makes  no  difference 
whether  it  is  measured  per  unit  of  original  or  per  unit  of  stretched 
length.  The  original  length  will  be  used  here. 

The  new  length  of  the  prism  is  l(i+X)  and  the  cross-section 
is    S(i—  X-^m)2.       The    volume     has     changed     from     SI     to 
.  5/(i+A  —  2&+m)  nearly,  if    higher  powers  of  A 

#1  than  the  first  are  dropped,  since  the  unit  deforma- 

tions are  very  small.      The  change  of  unit  volume 


r 

!   £  ,/  2 


is  therefore  A(  IT— ).     Thus,  if  w  is  nearly  4, 

for  metals,  the  change  of  volume  of  one  cubic 
unit  is   JA  nearly,  the  volume  being   increased 
for  longitudinal  tension.     If  there  were  no  change 
of  volume,  m  would  be  2,  as  is  the  case  for  india 
rubber  for  small  deformations.     Similarly,  for  compression  the 
change  of  the  unit  volume  is  nearly— JA  for  metals,  the  volume 
being  diminished. 

Example. — Steel,  £=29,000,000;  p=  20,000  Ib.  per  sq.  in.  tension; 

the  extension  will  be of  its  initial  length,  the  lateral  contraction 

i,45° 

will  be  about of  its  initial  width,  and  its  increase  of  volume  about 

5,800 

i 
2,900' 


INTERNAL  STRESS:    CHANGE  OF  FORM. 


185 


171.  Effect  of  Two  Principal  Stresses.  —  Denote  the  stresses 
by  pi  and  p2,  treated  as  tensile.  If  they  are  compressive,  reverse 
the  signs. 

Under  the  action  of  p1  there  will  be  the  following  stretch 
of  the  sides  per  unit  of  length  :  Fig.  76, 

parallel  to  O  C,  ~\ 

parallel  to  OB  and  O  A,  ~^ 
Under  the  action  of  p2  there  will  be 
parallel  to  O  B, 
parallel  to  O  A  and  O  C,  -  —  -p. 


|r; 


Adding  the  parallel  changes  or  stretches 


parallel  to  O  B,  A2=-^(#2-—  )  ; 


EV*    m 


parallel  to  O  A,  A3=  -— ^,(pi 


If  pi  and  p2  are  equal  unit  stresses,  but  of  opposite  signs, 
the  changes  of  length  become 


andzer°; 


or,  putting  either  of  these  two  changes  equal  to  A,  the  lengths 
of  the  sides  of  the  cube  originally  unity  per  edge  will  be  i+>l, 
i  —X,  and  i,  and  the  volume,  neglecting  A2,  is  unchanged. 


1  86  STRUCTURAL  MECHANICS. 

172.  Effect  of  Two  Shears.  —  The  cube  shown  in  Fig.  77 
has  been  deformed  by  the  action  of  two  equal  and  opposite  prin- 
cipal stresses,  and  a  square,  traced  on  one  side  of  the  original 
cube,  has  been  distorted  to  a  rhombus,  the  angles  of  which  are 
greater  and  less  than  a  right  angle  by  the  amount,  <j>.  By  §  160, 
two  equal  principal  stresses  of  opposite  sign  are  equivalent  to 
two  unit  shears  of  the  same  amount  per  square  inch  on 
planes  at  45°  with  the  principal  axes;  hence  the  distortion  of 
the  prism,  whose  face  is  the  rhombus,  results  from  two  equal 
shears  at  right  angles. 

Now  one-half  the  angle  JTT  —  <j>  has  for  its  tangent  J(i  —X)  -f- 
;  hence 

i  —  A  ,      i  -tan 


.  I+tani  . 

But  as  <f>  is  small,  A  =  J<£,  or  $  =  2\. 

Therefore  a  stretch  and  an  equal  shortening,  along  a  pair  of 

rectangular  axes,  are  equivalent  to  a  simple  distortion  relatively 

to  a  pair  of  axes  making  angles 
of  45°  with  the  original  axes  ;  and 
the  amount  of  distortion  is  double 
that  of  either  of  the  direct  changes 
of  length  which  compose  it.  This 
fact  also  appears  from  the  consid- 
eration that  a  distortion  of  a 
square  is  equivalent  to  an  elon- 
gation of  one  diagonal  and  a 

shortening  of  the  other  in  equal  proportions. 

Example.  —  For  steel,  as  before,  A=  —  —  ,  (f>  =  —  -  =  4'  45",  if  p\  — 
—  p2=  20,000  lb. 

173.  Modulus  of  Shearing  Elasticity.  —  Similarly,  equal  shear- 
ing stresses  q  on  two  pairs  of  faces  of  a  cube,  in  directions  par- 
allel to  the  third  face,  will  distort  that  third  face  into  a  rhombus, 
each  angle  being  altered  an  amount  <£,  there  being  distortion  of 
shape  only,  and  not  change  of  volume,  Fig.  78. 


INTERNAL  STRESS:    CHANGE  OF  FORM.  187 

Under  the  law  which  has  been  proved  true  within  the  elastic 
limit,  and  the  definition  of  the  modulus  of 
elasticity,  §  10,  a  modulus  of  transverse  (or 
shearing)  elasticity,  C,  also  called  coefficient 
of  rigidity,  as  E  may  be  called  coefficient  of 
stiffness,  may  be  written,  C  =  q-^<f>. 

As  these  two  unit  shears  are  equivalent  to 


-2 

a  unit  pull  and  thrust  of  the  same  magnitude  Fig.  78 

per   square  inch,   at   right   angles   with   each 
other  and  at  45°  with  these  shears,  the  case  is  identical  with  the 
preceding  one.     Then 

,     r  ,        -  .  -r  m  +  i 

and      J-t;  .       ... 


p     i     mE 
But,  as  p  =  q  =  C$,        c=-,=-- . 

<j>     2    m  + 1 

For  iron  and  steel  m  is  nearly  4,  which  gives  C  =  \E.  For 
wrought  iron  and  steel,  C  is  one  or  two  one-hundredths  less  than 
0.4-E.  Some  use  |£.  C=  11,000,000  is  a  fair  value. 

174.  Stress  on  One  Plane  the  Cause  of  Other  Stresses. — The 

elongation  produced  by  a  pull,  the  shortening  produced  by  a 
thrust,  and  the  distortion  due  to  a  shear  can  be  laid  off  as  graphi- 
cal quantities  and  discussed  as  were  unit  stresses  themselves.  All 
the  deductions  as  to  stresses  have  their  counterparts  in  regard  to 
changes  of  form.  There  has  been  found  an  ellipse  of  stress  for 
forces  in  one  plane,  when  two  stresses  are  given.  Also,  when 
three  stresses  not  in  one  plane  are  given,  there  is  an  ellipsoid  of 
stress  which  includes  all  possible  unit  stresses  that  can  act  on 
planes  in  different  directions  through  any  point  in  a  body.  So 
there  is  an  ellipse  or  ellipsoid  that  governs  change  of  form. 

Whether  the  movement  of  one  particle  towards,  from  or  by 
its  neighbor  sets  up  a  resisting  thrust,  pull  or  shear,  or  the  appli- 
cation of  a  pressure,  tension,  or  shear  is  considered  to  cause  a 
corresponding  compression,  extension,  or  distortion,  the  stresses 
and  the  elastic  change  of  form  coexist.  Hence  it  follows  that, 


i88  STRUCTURAL  MECHANICS. 

when  a  bar  is  extended  under  a  pull  and  is  diminished  in  lateral 
dimensions,  a  compressive  stress  acting  at  right  angles  to  the  pull 
must  be  aroused  between  the  particles,  and  measured  per  unit  of 
area  of  longitudinal  planes,  together  with  shears  on  some  inclined 
sections. 

That  such  a  state  of  things  can  exist  may  be  seen  from  the 
following  suggestions.     It  may  be  conceived  that  the  particles  of 
a  body  are  not  in  absolute  contact,  but  are  in  a 
state  of  equilibrium  from  mutual  actions  on  one 
another.     They  resist  with  increasing  stress  all 
attempts  to  make  them  approach  or  recede  from 
each  other,  and,  if  the  elastic  limit  has  not  been 
exceeded,  they  return  to  their  normal  positions 
when  the  external  forces  cease  to  act.     The  par- 
ticles in  a  body  under  no  stress  may  then  be  con- 
ceived to  be  equidistant  from  each  other.     The 
smallest  applied  external  force  will  probably  cause  change  in  their 
positions. 

If,  in  the  bar  to  which  tension  is  to  be  applied,  a  circle  is 
drawn  about  any  point,  experiment  and  what  has  been  stated 
about  change  of  form  in  different  directions  show,  that  the  diam- 
eter in  the  direction  of  the  pull  will  be  lengthened  when  the  force 
is  applied,  the  diameter  at  right  angles  will  be  shortened,  and  the 
circle  will  become  an  ellipse.  In  Fig.  79,  particle  i  moves  to  i', 
2  to  2',  4  to  4',  and  7  to  f.  As  they  were  all  equidistant  from 
o  in  the  beginning,  i  in  moving  to  if  offers  a  tensile  resistance, 
7  resists  the  tendency  to  approach  o,  while  a  particle  near  4,  mov- 
ing to  4',  does  not  change  its  distance  from  o,  but  moves  laterally 
setting  up  a  shearing  stress.  A  sphere  will  similarly  become  an 
ellipsoid. 

175.  Equivalent  Simple  Stress. — Our  knowledge  of  the  strength 
of  materials  is  derived  largely  from  experiments  upon  test  pieces 
in  which  the  load  is  applied  in  a  direction  along  the  piece — that  is, 
from  pieces  in  a  condition  of  simple  stress — and,  likewise,  working 
stresses  are  generally  considered  as  simple  stresses.  The  question 
then  arises  whether  a  body  which  is  stressed  in  two  directions — that 
is,  a  body  in  a  condition  of  compound  stress  from  the  action  of 


INTERNAL  STRESS:  CHANGE  OF  FORM.  189 

two  principal  stresses — is  stronger  or  weaker  than  if  stressed  in 
one  direction  only. 

For  example,  consider  the  stresses  acting  in  the  shell  of  a 
boiler  of  radius  r  and  steam  pressure  p.  §§  192  and  194  show 
that  on  any  longitudinal  section  there  is  a  tension  of  pr  per  unit 
of  length,  and  on  a  circumferential  section  there  is  a  tension  of 
\pr  per  unit  of  length.  The  tension  pr  produces  an  elongation  in 

the  direction  of  the  circumference  of  —  per  unit  of  length,  if  /  is 

the  thickness  of  the  plate.  But  by  §  170  the  tension  of  \pr  acting 
at  right  angles  to  pr  causes  the  plate  to  shorten  in  a  circumferential 

direction  an  amount  -  •  ~r  per  unit  of  length,  if  m  is  4  for  metals. 
The  resulting  lengthening  of  the  plate  circumferentially  under 
the  influence  of  the  two  principal  stresses  is,  therefore,  but  -— 

per  unit  of  length.  The  simple  stress  on  unit  length  of  plate  which 
would  produce  this  stretch  is  %pr,  and  if  the  deformation  of  the 
material  is  considered  to  be  a  measure  of  its  strength,  the  boiler 
is  stronger  because  of  the  stress  on  a  circumferential  joint  than 
it  would  be  if  such  stress  did  not  exist. 

Hence,  if  three  principal  stresses,  pi,-  p%,  ps,  exist  at  a  point 
in  a  body,  by  §  171  the  equivalent  simple  stress  which  would 
produce  the  same  deformation  at  that  point  in  the  direction  in 
which  pi  acts  is 


If  there  are  but  two  principal  stresses  as  in  most  of  the  problems 
discussed  in  this  chapter,  p3  is  zero.  If  p2  and  p3  are  of  opposite 
kind  to  pi,  the  equivalent  stress  pi  is  greater  than  pi. 

An  application  of  equivalent  stress  to  design  is  given  in  §  86. 
The  stresses  in  plates  of  §§  220  and  221  are  equivalent  stresses. 

Example. — At    a    certain   point    in  a    conical  steel  piston   there 


STRUCTURAL   MECHANICS. 

exist  principal  stresses  of  3,160  and  1,570  lb.,  of  opposite  signs.     Then 
/>/=3, 160+1-1,570=3,550  lb.;  p2f=  i, 570  +  1-3,160=  2,360  lb. 

176.  Cooper's  Lines. — Steel  plate  as  it  comes  from  the  mill 
has  a  firmly  adhering  but  very  brittle  film  of  oxide  of  iron  on  the 
surface.  This  film  is  dislodged  by  the  extension  of  a  test  speci- 
men in  tension  when  the  yield-point  is  passed.  Tf  a  hole  is 
punched  at  moderate  speed  in  a  steel  plate,  so  that  the  particles 
under  the  punch  have  some  opportunity  to  flow  laterally  under 
the  compression,  there  will  be  a  radial  compressive  stress  in  all 
directions  outwardly  from  the  circumference  of  the  hole,  and  a 
tensile  stress  circumferentially.  These  opposite  principal  stresses 
cause  shearing  planes  to  exist  whose  obliquity  depends  upon  the 
relative  magnitudes  of  the  principal  stresses  by  §  163.  The 
scale  breaks  on  these  lines  of  shear  and  there  result  curves  where 
the  bright  metal  shows  through,  branching  out  from  the  hole, 
intersecting  and  fading  away.  The  process  of  shearing  a  bar 
will  develop  the  same  curves  from  the  flow  of  the  metal  on  the 
face  at  the  cut  end.  They  are  known  as  Cooper's  lines.  These 
lines  show  that  deformation  takes  place  at  considerable  distances 
from  the  immediate  point  of  shearing  or  punching. 

Examples. — i.  A  pull  of  1,000  lb.  per  sq.  in.  and  a  thrust  of  2,000 
lb.  per  sq.  in.  are  principal  stresses.  Find  the  kind,  direction,  and 
magnitude  of  the  stress  on  a  plane  at  45°  with  either  principal  plane. 

2.  Find  the  stress  per  running  unit  of  length  of  joint  for  a  spiral 
riveted  pipe  when  the  line  of  rivets  makes  an  angle  of  45°  with  the 
axis  of  the  pipe  and  when  it  makes  an  angle  of  60°.  0.707  />;  0.5  p. 

3«.  A  rivet  is  under  the  action  of  a  shearing  stress  of  8,000  lb.  per 
sq.  in.  and  a  tensile  stress,  due  to  the  contraction  of  the  rivet  in  cooling, 
of  6,000  lb.  per  sq.  in.  Find  pi  and  p2. 

Pi  =  — 11,540  lb.;   />2=  +  5>54o  lb. 

4.  A  connecting  plate  to  which  several  members  are  attached  has 
a  unit  tension  on  a  certain  section  of  6,500  lb.  at  an  angle  of  30°  with 
the  normal.     On  a  plane  at  60°  with  the  first  plane  the  unit  stress  of 
5,000  lb.  compression  is  found  at  45°  with  its  plane.     Find  the  principal 
unit  stresses  and  the  shear.  —6,600;  +4,800;  5,700. 

5.  Assuming  the  weight  of  earth  to  be  105  lb.  per  cu.  ft.  and  the 
horizontal  pressure  to  be  one-third  the  vertical,  what  is  the  direction 
and  unit  pressure  per  square  foot  on  a  plane  making  an  angle  of  15° 
with  the  vertical  at  a  point  12  ft.  under  ground,  if  the  surface  is  level? 

515  lb.;  39^°  with  the  horizon. 

6.  A  stand-pipe,  25  ft.  diam.,  100  ft.  high.     The  tension  in  lowest 
ring,  if  }  in.  thick,  is  7,440  lb.  per  sq.  in.     If  plates  range  regularly 
from  J  in.  thick  at  base  to  \  in.  at  top,  neglecting  lap,  the  compression 
at  base  will  be  about  215  lb.  per  sq.  in.     For  a  wind  pressure  of  40 


INTERNAL  STRESS:    CHANGE  OF  FORM.  19 1 

lb.  per  sq.  ft.,  reduced  50%  for  cylindrical  surface,  and  treated  as  if 
acting  on  a  vertical  section,  M  at  base  =2, 500,000  ft.-lb.  Compression 
on  leeward  side  at  base  =  485  lb.  per  sq.  in.  If  ^  =  —  7,440  lb.,  p2= 
215  +  485  =  700  lb.,  find  the  stress  and  its  inclination  for  a  plane  at  30° 
to  the  vertical? 

Prove  that  the  shearing  plane  is  17°  04'  from  horizontal,  and  that 
the  shear  ic  2,284  lb-  per  sq.  in. 


CHAPTER  XII. 
RIVETS:    PINS. 

177.  Riveted  Joints. — There  are  four  different  ways  in  which 
riveted  joints  and  connections  may  fail.     The  rivets  may  shear 
off;    the  hole  may  elongate  and  the  plate  cripple  in  the  line  of 
stress;   the  plate  may  tear  along  a  series  of  rivet-holes,  more  or 
lees  at  right  angles  to  the  line  of  stress ;  or  the  metal  may  fracture 
between  the  rivet -hole  and  the  edge  of  the  plate  in  the  line  of 
stress.     From  the  consideration  that  a  perfect  joint  is  one  offering 
equal  resistance  to  each  of  these  modes  of  failure,  the  proper  pro- 
portions for  the  various  riveted  connections  are  deduced. 

178.  Resistance  to  Shear. — The  safe  resistance  of  a  rivet  to 
shearing  off  depends  upon  the  safe  unit  shear  and  the  area  of  the 
rivet  cross-section,  which  varies  as  the  square  of  the  diameter  of 
the  rivet.     When  one  plate  is  drawn  out  from  between  two  others, 
a  rivet  is  sheared  at  two  cross-sections  at  once,  and  is  twice  as 
effective  in  resisting  any  such  action.     Rivets  so  circumstanced 
are  said  to  be  in  double  shear,  and  their  number  is  determined 
on  that  basis. 

179.  Bearing  Resistance. — The  resistance  against  elongation 
of  the  hole  or  crippling  the  plate  depends  on  the  safe  unit  com- 
pression and  what  is  known  as  the  bearing  area,  the  thickness  of 
the  plate  multiplied  by  the  semicircumference  of  the  hole.     As 
the  semicircumference  varies  as  the   diameter,  it  is  more   con- 
venient, and  sufficiently  accurate,  to  use  the  product  of  the  thick- 
ness of  the  plate  and  the  diameter  of  the  rivet  with  a  value  of 
allowable   unit  compression   about   fifty  per  cent  greater   than 
usual.     In  practice  the  bearing  value  is  always  given  in  terms  of 
the  diameter. 

180.  Resistance  of  Plate. — The  resistance  to  tearing  across 

the  plate  through  a  line  of  holes,  or  in  a  zigzag  through  two  lines 

192 


RIVETS:    PINS.  193 

of  holes  in  the  same  approximate  direction,  depends  on  the  safe 
unit  tensile  stress  multiplied  by  the  cross-section  of  the  plate 
after  deducting  the  holes.  If  the  transverse  pitch,  or  distance 
between  centres  of  rivets,  is  considerable,  an  assumption  of 
uniform  distribution  of  tension  on  that  cross-section  is  not  likely 
to  be  true. 

The  resistance  of  the  metal  between  the  rivet-hole  and  the 
edge  of  the  plate  in  the  line  of  stress  is  usually  taken  as  the  safe 
unit  shear  for  the  plate  multiplied  by  the  thickness  and  twice 
the  distance  from  the  rivet- hole  to  the  edge.  Some,  however, 
consider  that  the  resisting  moment  of  the  strip  of  metal  in  front 
of  the  rivet-holes  is  called  into  action. 

181.  Bending:  Friction. — There  are  those  who  advise  the  com- 
puting of  a  rivet  shank  as  if  it  were  subjected  to  a  bending  moment. 
If  the  rivet  fills  the  hole  and  is  well  driven,  there  is  no  bending 
moment  exerted  on  it,  unless  it  passes  through  several  plates. 
As  practical  tests  have  shown  that  rivets  cannot  surely  be  made 
to  fill  the  holes,  if  the  combined  thickness  of  the  plates  exceeds 
five  diameters  of  the  rivet,  this  limitation  will  diminish  the  im- 
portance of  the  question  of  bending. 

No  account  is  taken  of  the  friction  induced  in  the  joint  by 
its  compression  and  the  cooling  of  the  rivet,  and  such  friction 
gives  added  strength.  As  the  rivet  is  closed  up  hot,  the  shank 
is  under  more  or  less  tension  when  cold.  Moreover,  the  head 
is  not  given  the  thickness  required  in  the  head  of  a  bolt  under 
tension,  and  therefore  rivets  are  not  available  for  any  more  ten- 
rion,  and  should  not  be  used  for  that  purpose.  Tight-fitting 
turned  bolts  are  required  in  such  a  case. 

182.  Spacing. — The  rivets  should  be  well  placed  in  a  joint 
or  connection,  in  order  to  insure  a  nearly  uniform  distribution  of 
stress  in  the  piece;    they  should  be  symmetrically  arranged,  be 
placed  where  they  can  be  conveniently  driven,  and  be  spaced 
so  that  the  holes  can  be  definitely  and  easily  located  in  laying 
out  the  work. 

183.  Minimum  Diameter  of  Rivets. — The  punch  must  have 
a  little  clearance  in  the  die.     The  wad  of  metal  shears  out  below 
the  punch  with  more  ease  and  with  less  effect  on  the  surrounding 


1 94 


STRUCTURAL  MECHANICS. 


metal  when  it  can  flow,  as  it  were,  a  little  laterally,  and  it  then 
comes  out  as  a  smooth  frustum  of  a  cone  with  hollowed  sides, 
reminding  one  of  the  vena  contracta.  The  punch  must  also  be  a 
little  larger  than  the  rivet,  to  permit  the  ready  entrance  of  the 
rivet-shank  at  a  high  heat.  The  diameter  of  the  hole  is  com- 
monly computed  at  J  inch  in  excess  of  the  nominal  diameter 
of  the  rivet;  but  the  rivet  is  treated  as  if  of  its  nominal  diameter. 

One  other  consideration  has  weight  in  determining  the  minimum 
diameter  of  the  rivet.  If  the  diameter  of  the  rivet  is  less  than 
the  thickness  of  the  plate,  the  punch  will  not  be  likely  to  endure 
the  work  of  punching.  A  diameter  one  and  a  half  times  the 
thickness  of  the  plate  is  often  thought  desirable. 

184.  Number  and  Size  of  Rivets. — Formulas  are  of  little  or 
no  value  in  designing  ordinary  joints  and  connections.  Boiler 
joints  and  similar  work  can  be  computed  by  formulas,  but  to  no 
great  advantage.  Tables  are  used  which  give  what  is  termed 
the  shearing  value  of  different  rivet  cross-sections  in  pounds 
for  a  certain  allowable  unit  shear,  and  the  bearing  or  compression 
value  of  different  thicknesses  of  plate  and  diameters  of  rivet  for 
a  certain  allowable  unit  compression.  For  a  given  thickness 
of  plate,  that  diameter  of  rivet  is  the  best  whose  two  values,  as 
above,  most  nearly  agree.  The  quotient  of  the  force  to  be  trans- 
mitted through  the  connection  or  through  a  running  foot  of  a 
boiler  joint,  divided  by  the  less  of  the  two  practicable  values,  will 
give  the  minimum  number  of  rivets.  Their  distribution  is  governed 
by  the  considerations  previously  referred  to.  Whether  a  joint 
in  a  boiler  requires  one,  two,  or  three  rows  of  rivets  depends  upon 
the  number  needed  per  foot. 

Example. — Two  tension  bars,  6  in.  by  J  in.,  carrying  30,000  lb., 
are  to  be  connected  by  a  short  plate  on  each  side.  Let  unit  shear  be 
7,500  lb.  per  sq.,  in.,  unit  bearing  15,000  lb.  on  the  diameter,  and  unit 
tension  12,000  lb.  The  bearing  value  of  a  f-in.  rivet  in  a  £-in.  plate 
is  5,625  lb.,  its  shearing  value  in  double  shear  is  2X3,310=6,620  lb. 
Hence  30,000-^5,625  =  6  rivets  necessary.  If  these  rivets  can  be  so 
arranged  that  a  deduction  of  but  one  rivet-hole  is  necessary  from  the 
cross-section  of  the  tie,  (6—  (|+J))i=2.56  sq.  in.  net  section,  which 
will  carry  30,750  lb.  at  12,000  lb.  unit  tension.  Each  cover-plate  can- 
not be  less  than  }  in.  thick,  and,  as  will  be  seen  presently,  should  be 


RIVETS:    PINS. 


made  a  little  more.     The  length  will  depend  on  the  distribution  of  the 
rivets,  to  be  taken  up  next. 

185.  Arrangement    of   Rivets. — Long    joints,    under    tension, 
like  those  of  boilers,  are  connected  by  one  or  more  rows  of  rivets, 


0 

D 

o 

0 

o 

o 

o 

o  o  o 
o  o  o 


000 


o     o 


1 
1 

^J 

r 

vr 

U 

f  ~^U    °  O 

G     I 

1 

!o 

0 
C 

1 
)iO 

o 

0 

(o     o  oo 

to    I 

1 

k 

JX^  O    O  O 

n 

m 

1 

1  1 
if  \ 

\\ 

S    c 

[ 

0  0 
0 

0 

o 

o  o||o  o  o  o\ 
o      H  o  o  o  c 

\ 
D| 

Li 

o 

o 

O             O  O  O  ( 

J  . 

!     ! 

I 

o  o 

o 

o  op  o  o  o/ 

/ 

Fig.  80 


as  shown  at  A  and  B,  Fig.  80.  If  more  than  one  row  is  needed, 
the  rivets  are  staggered,  as  at  B,  and  the  rows  should  be  separated 
such  a  distance  that  fracture  by  tension  is  no  more  likely  on  a 


196  STRUCTURAL  MECHANICS. 

zigzag  line  than  across  a  row.  Experiments  have  shown  that  a 
plate  will  break  along  a  zigzag  line  such  as  is  shown  at  B  unless 
the  net  length  of  that  line  exceeds  by  about  one-third  the  net 
length  of  a  line  through  a  row.  To  prevent  tearing  out  at  the 
edge  of  the  plate,  the  usual  specification  of  at  least  one  and  a 
half  rivet  diameters  from  centre  of  hole  to  edge  of  plate  will 
suffice. 

The  tendency  of  a  lap-joint  to  cause  an  uneven  distribution 
of  stress  by  reason  of  bending,  and  the  same  tendency  when  a 
single  cover  is  used,  is  shown  at  C.  The  increase  of  stress  thus 
caused  should  be  offset  by  increased  thickness  of  plate.  A  cover 
strip  on  each  side  is  preferable  if  not  objectionable  for  other 
reasons. 

In  splicing  ties,  D  shows  a  bad  arrangement,  the  upper  plan 
failing  to  distribute  the  stress  evenly  across  the  tie,  and  the  lower 
plan  wasting  the  section  by  excessive  cutting  away.  The  rivets 
at  E  are  well  distributed  across  the  breadth,  and  weaken  the  tie 
by  but  one  hole,  as  only  two-thirds  of  the  stress  passes  the  section 
reduced  by  two  holes;  and,  unless  the  net  section  at  this  place 
is  less  than  two-thirds  of  the  section  reduced  by  one  hole,  it  is 
equally  strong.  Thus  b-d=%(b — 2$,  or  a  breadth  equal  to 
or  greater  than  four  diameters  will  satisfy  this  requirement. 
The  covers,  however,  will  be  weakened  by  two  holes,  and  hence 
their  combined  thickness,  when  two  are  used,  should  exceed  the 
thickness  of  the  tie. 

F  similarly  is  better  than  G,  and  the  tie  at  F  is  again  weakened 
by  but  one  hole.  The  sectional  area  of  the  plate  shown  at  H 
is  diminished  by  two  holes  at  m  and  four  at  n,  but  the  stress 
on  section  n  is  less  than  that  on  m  by  the  stress  the  rivets  at  m 
transmit  to  the  splice  plate.  Consequently,  in  designing  a  splice, 
if  the  area  cut  out  by  the  two  extra  rivets  at  n  multiplied  by  the 
working  stress  in  the  plate  does  not  exceed  the  working  value 
of  the  two  rivets  at  m,  the  plate  will  be  weakened  by  two  holes 
only.  In  the  splice  shown  the  sections  at  m  and  n  will  be  equally 
strong  when  b-2d={^(b-<\d)  or  when  b  =  iSd. 

As  it  is  desirable  to  transmit  all  but  the  proper  fraction  of  the 
tension  past  the  first  rivet,  the  corners  of  the  cover  F  or  H  are 


RIVETS:    PINS.  1 97 

clipped  off,  thus  increasing  the  unit  tension  in  the  reduced  section 
and  increasing  its  stretch  to  correspond  more  nearly  with  the 
unit  tension  and  elongation  of  the  tie  beneath.  The  appearance 
of  the  connection  is  also  improved. 

It  is  not  desirable  to  make  splice  plates  as  short  as  possible, 
because  a  short  splice  is  likely  to  be  weak.  In  splicing  members 
built  up  of  shapes  this  is  especially  the  case,  as  a  uniform  dis- 
tribution of  stress  over  the  whole  cross-section  is  not  easily  secured 
when  the  stress  passes  into  the  member  within  a  length  less 
than  its  width.  Short  splice  plates  may  be  lengthened  without 
increasing  the  number  of  rivets  by  using  a  greater  pitch. 

186.  Remarks. — If  the  member  is  in  compression,  the  holes 
are  not  deducted,  since  the  rivets  completely  fill  the  holes;  and 
the  strength  is  computed  on  the  gross  section.  Unless  special 
care  is  exercised  in  bringing  two  connected  compression  pieces 
into  close  contact  at  their  ends,  good  practice  requires  the  use  of 
a  sufficient  number  of  rivets  at  the  connection  to  transmit  the 
given  force. 

Rivet-heads  in  boiler  work  are  flat  cones.  In  bridge  and 
structural  work  they  are  segments  of  spheres,  known  as  button 
heads,  and  are  finished  neatly  by  means  of  a  die.  These  heads 
may  be  flattened  when  room  is  wanting,  and  countersunk  heads 
are  used  where  it  is  necessary  to  have  a  finished  flat  surface. 

Members  of  a  truss  which  meet  at  an  angle  are  connected 
by  plates  and  rivets.  The  axes  of  the  several  members  should 
if  possible  intersect  in  a  common  point.  If  they  do  not,  moments 
are  introduced  which  give  rise  to  what  are  known  as  secondary 
stresses,  as  distinguished  from  the  primary  stresses  due  to  the 
direct  forces  in  the  pieces  of  the  frame.  Such  secondary  stresses 
may  be  of  considerable  magnitude  in  an  ill-designed  joint. 

It  is  desirable  to  arrange  the  rivets  in  rows  which  can  be 
easily  laid  out  in  the  shop,  and  to  make  the  spacing  regular, 
avoiding  the  use  of  awkward  fractions  as  much  as  possible. 

Commercial  rivet  diameters  vary  by  eighths  of  an  inch,  f-, 
f-,  and  {-in.  rivets  being  the  ones  frequently  used.  As  much 
uniformity  as  possible  in  the  size  of  rivets  will  tend  to  economy 
in  cost. 


IQ8  STRUCTURAL  MECHANICS. 

187.  Structural  Riveting. — The  following  rules  for  structural 
work  are  in  harmony  with  good  practice: 

Holes  in  steel  f  in.  thick  or  less  may  be  punched;  when 
steel  of  greater  thickness  is  used,  the  holes  shall  be  subpunched 
and  reamed  or  drilled  from  the  solid. 

The  diameter  of  the  die  shall  not  exceed  that  of  the  punch 
by  more  than  &  of  an  inch,  and  all  rivet-holes  shall  be  so  accu- 
rately spaced  and  punched  that,  when  the  several  parts  are 
assembled  together,  a  rivet  &  in.  less  in  diameter  than  the  hole 
can  generally  be  entered  hot  into  any  hole. 

The  pitch  of  rivets,  in  the  direction  of  the  stress,  shall  never 
exceed  6  in.,  nor  16  times  the  thickness  of  the  thinnest  plate 
connected,  and  not  more  than  30  times  that  thickness  at  right 
angles  to  the  stress. 

At  the  ends  of  built  compression  members  the  pitch  shall  not 
exceed  4  diameters  of  the  rivet  for  a  length  equal  to  twice  the 
width  of  the  member. 

The  distance  from  the  edge  of  any  piece  to  the  centre  of  a 
rivet-hole  must  not  be  less  than  ij  times  the  diameter  of  the 
rivet,  nor  exceed  8  times  the  thickness  of  the  plate;  and  the 
distance  between  centres  of  rivet-holes  shall  not  be  less  than  3 
diameters  of  the  rivet. 

The  effective  diameter  of  a  driven  rivet  will  be  assumed  to  be 
the  same  as  its  diameter  before  driving;  but  the  rivet- hole  will 
be  assumed  to  be  one-eighth  inch  diameter  greater  than  the  un- 
driven  rivet. 

In  structural  riveting  these  relationships  between  unit  working 
stresses  are  very  commonly  used: 

Bearing  stress  =  i  J  X  tensile  stress ; 
Shearing  stress  =  £  X bearing  stress  =  f  X tensile  stress. 
The  shearing  area  of  the  rivets,  therefore,  should  exceed  by 
one-third  the  net  area  of  the  tension  member  they  connect.     See 
§§  146,  147,  148. 

188.  Boiler-riveting. — Boiler  work  admits  of  standardization 
much   more  readily  than  structural  work  and  standard   boiler 
joints,  which  make  the  tensile  strength  of  the  net  plate  equal  tu 
the  strength  of  the  rivets,  have  been  very  generally  adopted.    A 
triple-riveted  boiler  joint  is  shown  in    Fig.  80,  L.      The  most 
notable  point  of  difference  between  boiler  and  structural  riveting 
is  that  it  is  not  customary  to  consider  the  bearing  of  rivets  in  boiler 


RIVETS:    PINS.  199 

work.  Rivets  are  figured  for  shear  only  and  in  double  shear  are 
considered  to  have  but  one  and  three-quarters  time  the  value  of 
rivets  in  single  shear,  instead  of  twice  the  value,  as  in  structural 
work.  Custom  in  this  country  makes  the  allowable  unit  shear 
on  rivets  approximately  two-thirds  the  unit  tensile  stress  in  the 
plate,  but  the  British  Board  of  Trade  rule  gives  about  four-fifths. 

The  diameter  of  the  rivets  used  should  be  about  twice  the  thick- 
ness of  the  plate.  In  ordinary  cases  there  is  no  danger  that  the 
rivets  will  be  too  far  apart  to  render  the  joint  water-  or  steam- 
tight,  when  the  edge  of  the  plate  is  properly  closed  down  with  a 
calking-tool. 

189.  Pins:  Reinforcing  Plates. — The  pieces  of  a  frame  are 
frequently  connected  by  pins  instead  of  rivets.  The  axes  of  the 
several  pieces  are  thus  made  to  meet  in  a  common  point,  if  the 
pin-hole  is  central  in  each  member.  Pins  are  subjected  to  com- 
pression on  their  cylindrical  surfaces,  to  shear  on  the  cross-section, 
and  to  bending  moments.  The  compression  on  the  pin-hole  is 
reduced  to  the  proper  unit  stress,  if  necessary,  by  riveting  re- 
inforcing plates  to  the  sides  of  the  members,  as  shown  at  K, 
Fig.  80.  A  sufficient  number  of  rivets  to  transmit  the  proper 
proportion  of  the  force  must  be  used,  with  a  due  consideration 
of  the  shearing  value  of  a  rivet  and  its  bearing  value  in  the  re- 
inforcing plate  or  the  member  itself,  whichever  gives  the  less  value. 
No  more  rivets  should  be  considered  as  efficient  behind  the  pin 
than  the  section  of  the  reinforcing  plate  each  side  of  the  pin-hole 
will  be  equivalent  to. 

When  the  pin  passes  through  the  web  of  a  large  built  member, 
such  as  a  post  or  a  top  chord  of  a  bridge,  the  web  is  often  so  thin 
that  more  than  one  reinforcing  plate  on  either  side  is  needed. 
It  is  then  economical  to  make  the  several  plates  of  increasing 
length,  the  shortest  on  the  outside,  and  determine  the  number 
of  rivets  in  each  portion  accordingly. 

Pin -plates  should  be  made  long  for  the  same  reason  as  given 
for  making  long  splice  plates.  The  longest  plate  is  sometimes 
required  to  extend  6  in.  inside  the  tie-plates  so  that  the  stress 
may  be  transferred  to  the  flanges  and  not  overtax  the  web.  See 
K,  Fig.  80. 


2OO 


STRUCTURAL  MECHANICS. 


190.  Shear  and  Bearing. — The  shear  at  any  section  of  the  pin 
is  found  from  the  given  forces  in  the  pieces  connected.     The 
resultant  of  the  forces  in  the  pieces  on  one  side  of  any  pin  section 
will  be  the  shear  at  that  section.     As  the  pin  will  probably  not 
fit  the  hole  tightly  (a  difference  of  diameter  of  one-fiftieth  of  an 
inch  being  usually  permitted),  the  maximum  unit  shear  will  be 
four- thirds  of  the  mean  (§  72).     Specifications  frequently  give  a 
reduced  value  for  mean  unit  shear,  which  provides  for  this  un- 
equal distribution.     Bearing  area  is  figured  as  if  projected  on 
the  diameter. 

191.  Bending  Moments  on  Pins. — At  a  joint  where  several 
pieces  are  assembled,  the  resisting  moment  required  to  balance 
the  maximum  bending  moment  on  the  pin  caused  by  the  forces 
in  those  pieces  will  generally  determine  the  diameter  of  the  pin. 
In  computing  the  bending  moments,  the  centre  line  of  each  piece 
or  bearing  is  considered  the  point  of  application  of  the  force 
which  it  carries.     This  assumption  is  likely  to  give  a  result  some- 
what in  excess  of  the  truth,  as  any  yielding  tends  to  diminish 
the  arm  of  each  force. 

The  process  of  finding  the  bending  moments  will  be  made 
clear  by  an  illustration.     Fig.  81  shows  the  plan  and  elevation 


I  <> 

^D               ^ 

B 

Fig.  81 

of  the  pieces  on  a  pin,  with  the  forces  and  directions  marked. 
The  thickness  of  the  pieces,  which  are  supposed  to  be  in  contact, 
is  also  shown.  The  joint  must  be  symmetrically  arranged,  to 
avoid  torsion,  and  simultaneous  forces  must  be  used,  which  reduce 


RIVETS:    PINS.  2OI 

to  zero  for  equilibrium.  As  the  joint  is  symmetrical,  the  com- 
putation is  carried  no  farther  than  the  piece  adjoining  the  middle. 
Resolve  the  given  forces  on  two  convenient  rectangular  axes, 
here  horizontal  and  vertical.  Set  the  horizontal  components  in 
order  in  the  column  marked  H,  the  vertical  ones  in  the  column 
marked  F.  Their  addition  in  succession  gives  the  shears,  marked 
F.  The  next  column  shows  the  distance  from  centre  to  centre 
of  each  piece.  Fdx  is  then  the  increment  of  bending  moment; 
and  the  summation  of  increments  gives,  in  the  column  M,  the 
bending  moment  at  the  middle  of  each  piece,  from  the  horizontal 
and  from  the  vertical  components  respectively.  The  square  root 
of  the  sum  of  the  squares  of  any  pair  of  component  bending 
moments  will  be  the  resultant  bending  moment  at  that  section. 
It  is  comparatively  easy  to  pick  out  the  pair  of  components  which 
will  give  a  maximum  bending  moment  on  the  pin.  Equate  this 
value  with  the  resisting  moment  of  a  circular  section  and  find 
the  necessary  diameter. 

H.  F.  dx.  Fdx.  M. 

A  + 10,000  + 10,000 


B 

—40,000 

—  30,000 

+  11,250 

+  11,250 

i 

C 

0 

—  30,000 

5 

-22,500 

-11,250 

D 

+  15,000 

-15,000 

8 

-18,750 

-30,000 

i 

E 

.    +15,000 

-!3.I25 

-43>I25 

o 

F. 

A 

-   5,780 

-   5,78o 

B 

-    5,780 

-  6,503 

-  6,503 

f 

C 

-   2,890 

-   8,670 

-  4,335 

-10,838 

f 

D 

-  8,670 

-  5,4i9 

-16,257 

I 

E 

+    8,670 

-  7,586 

-23343 

202  STRUCTURAL  MECHANICS. 

M  at  D=\/ (30,0002  +  16,2572) ;  M  at  E  =  v/ (43,1252 +  23,8432). 
The  latter  is  plainly  the  larger,  and  is  49,210  in.-lb. 

The  pieces  can  be  rearranged  on  this  pin  to  give  a  smaller 
moment.  The  maximum  moment  is  not  always  found  at  the 
middle. 

The  bending  moment  at  any  point  of  the  beam  or  shaft,  when 
the  forces  do  not  lie  in  one  plane,  can  be  found  in  the  same  way. 

A  solution  of  the  above  problem  by  graphics  can  be  found 
In  the  author's  Graphics,  Part  II,  Bridge  Trusses. 

Examples. — i.  A  tie-bar  J  in.  thick  and  carrying  24,000  Ib.  is 
spliced  with  a  butt-joint  and  two  covers.  If  unit  shear  is  7.500  Ib., 
unit  bearing  on  diameter  is  15,000  Ib.,  and  unit  tension  is  10,000  Ib., 
find  the  number,  pitch,  and  arrangement  of  J-in.  rivets  needed,  and 
the  width  of  the  bar. 

2.  The  longitudinal  lap-joint  of  a  boiler  must  resist  52,000  Ib.  ten- 
sion per  linear  foot.  If  the  unit  working  stress  for  the  shell  is  12,000 
Ib.  and  the  other  stresses  as  above,  \vhat  size  of  rivet  is  best,  for  triple 
riveting,  what  the  pitch,  and  the  thickness  of  the  shell? 


CHAPTER  XIIL 
ENVELOPES. 

192.  Stress  in  a  Thin  Cylinder. — Boilers,  tanks,  and  pipes 
under  uniform  internal  normal  pressure  of  p  per  square  inch. 

Conceive  a  thin  cylinder,  of  radius  r,  to  be  cut  by  any  diame- 
tral plane,  such  as  the  one  represented  in  Fig.  82,  and  consider 


Fig  82 


the  equilibrium  of  the  half  cylinder,  which  is  illustrated  on  the 
left.  It  is  evident  that,  for  unity  of  distance  along  the  cylinder, 
the  total  pressure  on  the  diameter,  2pr,  must  balance  the  sum 
of  the  components  of  the  pressure  on  the  semi-circumference  in 
a  direction  perpendicular  to  the  diameter.  This  pressure,  2pr, 
uniformly  distributed  over  the  diameter,  must  cause  a  tension 
T  in  the  material  at  each  end  to  hold  the  diameter  in  place.  Hence 

T-pr. 

As  all  points  of  the  circle  are  similarly  situated,  the  tension  in 
the  ring  at  all  points  is  constant  and  equal  to  pr.  If  the  thick- 
ness is  multiplied  by  the  safe  working  tension  /  per  square  inch, 
it  may  be  equated  with  pr,  giving 

Required  net  thickness  =pr+}. 

In  a  boiler  or  similar  cylinder  made  up  of  plates  an  increase 
of  thickness  will  be  required  to  compensate  for  the  rivet-holes. 
If  a  is  the  pitch,  or  distance  from  centre  to  centre,  of  consecu- 

203 


204  STRUCTURAL  MECHANICS. 

tive  rivets  in  one  row  along  a  joint,  and  d  the  diameter  of  the 
rivet-hole,  the  effective  length  a  to  carry  the  tension  is  reduced 
to  a—  d,  and  the  gross  thickness  of  plate  must  not  be  less  than 

pr     a 


Example. — The  circumferential  tension  in  a  boiler,  4  ft.  diameter, 
carrying  120  Ib.  steam  pressure  is  120-24=2,880  Ib.  per  linear  inch  of 

length  of  shell,  which  will  require  a  plate  —     -  in.  thick  (net),  if  /  is 

10,000 

not  to  exceed  10,000  Ib.  per  sq.  in.     Net  thickness =-^g-  in.     If  a  longi- 
tudinal joint   has  f-in.   rivet-holes,  at  2\  in.  pitch,  in  two  rows,  the 

2  880- 2^ 

thickness  of  plate  must  not  be  less  than  -         — ?*=  A  m- 

10,000 -if 

193.  Another  Proof  of  the  value  of  T  may  be  obtained  as 
follows:  The  small  force  on  arc  ds=pds.  The  vertical  compo- 
nent of  this  force  =  pds  sin  0  =  pdx.  The  entire  component  on 

r  +  r 
one  side  of  the  diameter  is  /      pdoc  =  2  i>r,  which  must  be  resisted 


/  +  r 
pdoc  =  2pr,  whi 


by  the  tension  in  the  matei  LI  at  the  two  ends  of  the  diameter. 

The  same  result  will  be  obtained  graphically  by  laying  off  a 
load  line  =  I  'pds,  which  becomes  a  regular  polygon  of  an  infinite 
number  of  sides,  i.e.,  a  circle,  with  the  lines  to  the  pole  making 
the  radii  of  the  length  pr. 

The  cylinder,  under  these  circumstances,  is  in  stable  equi- 
librium. If  not  perfectly  circular,  it  tends  to  become  so,  small 
bending  moments  arising  where  deviation  from  the  circle  exists. 
Hence  a  lap-joint  in  the  boiler  shell  causes  a  stress  from  the 
resisting  moment  to  be  combined  with  the  tension  at  the  joint. 

The  above  investigation  applies  only  to  cylinders  so  thin  that 
the  tension  may  be  considered  as  distributed  uniformly  over  the 
section  of  the  plate. 

For  riveting  see  Chapter  XII. 

194.  Stress  in  a  Right  Section.  —  The  total  pressure  from  p 
on  a  right  section  of  the  cylinder  is  nr2p,  which  will  also  be  the 
resultant  pressure  on  the  head  in  the  direction  of  the  axis  of  the 
cylinder,  whether  the  head  is  flat  or  not.  This  pressure  causes 
tension  in  every  longitudinal  element  of  the  cylinder,  or  in  every 
cross-section.  As  this  cross-section  is  2xrX  thickness,  the  longi- 
tudinal tension  per  linear  inch  of  a  circumferential  joint  is 


ENVELOPES.  205 

7tr2p  +  27tr  =  %pr,  or  one-half  the  amount  per  linear  inch  of  a 
longitudinal  joint.  Hence  a  boiler  is  twice  as  strong  against 
rupture  on  a  circumferential  joint  as  on  a  longitudinal  joint, 
and  hence  the  longitudinal  seams  are  often  double  riveted  while 
the  circumferential  ones  are  single  riveted. 

195.  Stress  in  any  Curved  Ring  under  Normal  Pressure.  — 
The  stress  in  a  circular  ring  of  radius  r,  under  internal  or  ex- 
ternal normal  unit  pressure,  p,  is  pr  per  linear  unit  of  section  of 
the  ring,  being  tension  in  the  first  case  and  compression  in  the 
second.     Similarly,    in   single   curved   envelopes    in   equilibrium 
under  normal  pressure  (that  is,  envelopes  in  which  the  stress  acts 
in  the  direction  of  the  shell)  the  stress  at  any  point  per  linear 
unit  along  an  element  is  equal  to  pp,  in  which  p  is  the  radius 
of  curvature  of  the  curve  cut  out  by  a  plane  normal  to  the  element 
and    passing    through    the    given 

point.  Fig.  83  shows  the  trace 
of  an  arc  of  shell  of  width  unity 
and  of  length  pdd  acted  upon  from 
within  by  a  normal  pressure,  p. 
Then  P  =  ppdO  and  for  equilibrium 
the  sum  of  the  components  acting 
in  the  direction  of  P  must  be  zero 
or  P  =  2  T  sin  \dB  =  Tdd.  Hence  T  =  pp. 

196.  Thin  Spherical  Shell  :  Segmental  Head.  —  If  a  thin  hollow 
sphere  of  radius  /  has  a  uniform  normal  unit  pressure  p  applied 
to  it  within,  the  total  interior  pressure  on  a  meridian  plane  will 
be  nr'2p,  and  the  tension  per  linear  inch  of  shell  will  be 


If  p  is  applied  externally,  the  stress  in  the  material  will  be  com- 
pression. It  may  be  noted  that  the  double  curvature  of  the 
sphere  is  associated  with  half  the  stress  which  is  found  in  the 
cylinder  of  single  curvature  having  the  same  radius. 

If  a  segment  of  a  sphere  is  used  to  close  or  cap  the  end  of 
a  cylinder  or  boiler,  the  same  value  will  hold  good.  In  this  case 
the  radius  r'  is  greater  than  r  for  the  cylinder. 

If  the  segmental  end  is  fastened  to  the  cylinder  by  a  bolted 


206  STRUCTURAL  MECHANICS. 

flange,  the  combined  tension  on  the  bolts  will  be  xr2p,  as  this 
is  the  total  force  on  a  right  section  of  the  cylinder. 

The  flange  itself  will  be  in  compression.  The  pressure  p 
from  below,  in  Fig.  84,  causes  a  pull  per  circumferential  unit 
in  the  direction  of  a  tangent  at  B,  which  pull  has  just  been  shown 
to  be  equal  to  %pr> '.  It  may  be  resolved  into  vertical  and  hori- 
zontal components.  The  vertical  component  B  C  is,  by  §  194, 
\pr.  The  horizontal  component  h  must  be  proportioned  to  the 
vertical  component  as  A  O  to  A  B,  the  sides  of  the  right-angled 

triangle  to  which  they  are   respec- 
c*      '^x^          rf/7>s^  tively    perpendicular.      As    AO  = 

V(r'2-r2), 

P~*\ 

As  h  is  a  uniform  normal  pressure 
applied   from   without    (or   tension 

applied  from  within)  in  the  plane  of  the  flange,  the  compression 
on  the  cross-section  of  the  latter  will  be  hr  or  ^pr\/(rf2  —  r2),  to 
be  divided  by  that  cross-section  for  finding  the  unit  compression. 
Segmental  bottoms  of  cylinders  are  sometimes  turned  inward. 
The  principles  are  the  same. 

Example. — A  segmental  spherical  top  to  a  cylinder  of  24  in.  diam- 
eter, under  100  Ib.  steam  pressure,  has  a  radius  of  15  in.  with  a  versed 
sine  of  6  in.  The  tension  in  top  =  £- 100- 15  =  750  Ib.  per  linear  inch. 
If  its  thickness  is  }  in.,  the  stress  per  sq.  in.  is  3,000  Ib.  The  total  pull 
on  the  flange  bolts  is  100-144-22-^7  =  45,260  Ib.  A  j-in.  bolt  has 
about  0.3  sq.  in.  section  at  bottom  of  thread,  giving  a  tension  value  of 
about  3,000  Ib.  if  /=  10,000  Ib.  There  would  be  needed  some  15  bolts, 
about  5J  in.  centre  to  centre  on  a  circumference  of  26  in.  diameter. 
The  compression  in  the  flange  is  £-100-12-9=5,400  Ib.  A  2X^-in. 
flange  with  a  f-in.  hole  has  a  section  J-i}=J  sq.  in.,  giving  a  unit 
compression  in  the  flange  of  1-5,400=8,600  Ib.  per  sq.  in. 

A  similar  compression  acts  in  the  connecting  circle  between  a 
water-tank  and  the  conical  or  spherical  bottom  sometimes  built.  See 
§§  204,  205. 

197.  Thick  Hollow  Cylinder. — If  the  walls  of  a  hollow  cylin- 
der or  sphere  are  comparatively  thick,  it  will  not  be  sufficiently 
accurate  to  assume  that  the  stress  in  any  section  is  uniformly 
distributed  throughout  it.  If  the  material  were  perfectly  rigid, 


ENVELOPES. 


207 


the  internal  or  external  pressure  would  be  resisted  by  the  imme- 
diate layer  against  which  the  pressure  was  exerted,  and  the  re- 
mainder of  the  material  would  be  useless.  As,  however,  the 
substance  of  which  the  wall  is  composed  yields  under  the  force 
applied,  the  pressure  is  transmitted  from  particle  to  particle, 
decreasing  as  it  is  transmitted,  since  each  layer  resists  or  neu- 
tralizes a  portion  of  the  normal  pressure  and  undergoes  ex- 
tension or  compression  in  so  doing. 

198.  Greater  Pressure  on  Inside. — Let  Fig.  85  represent  the 
right  section  of  a  thick  hollow  cylinder,  such  as  that  of  a  hydraulic 
press.  Let  r±  and  r2  be  the  internal  and  external  radii  in  inches; 
pi  and  p2  the  internal  and  external  normal  unit  pressures  in 
pounds  per  square  inch,  pi  being  the  greater;  and  p  the  unit 
normal  pressure  on  any  ring  whose  radius  is  r. 

If  a  hoop  is  shrunk  on  to  the  cylinder,  p2  will  be  the  unit 
normal  pressure  thus  applied  to  the  exterior  of  the  cylinder. 

The  unit  tensile  stress  found  in  a  thin  layer  of  radius  r  and 
thickness  dr  will  be  denoted  by  /,  and  will  be  due  to  that  portion 
0}  p  which  is  resisted  by  the  layer  and  not  transmitted  to  the  next 
exterior  layer.  The  total  tension  on  the  radial  section  of  a  ring 
lying  between  ri  and  r  is  piri—pr,  since  the  pressure,  pi,  on 
the  inside  sets  up  a  tension  of  $\r\  in  the  ring  and  the  pressure, 
p,  acting  on  the  outside  of  the  ring  sets  up  a  compression  of  pr. 
This  total  tension  may  also  be  expressed  by 

/tdr.     As   p  and  r  are  variables,  there  is 
obtained  by  differentiating  the  equation 

piri-pr=   I    tdr, 

Jr\ 


or 


-d(pr)=tdrt 
pdr+rdp+tdr=o. 


(i) 


Another  equation  can  be  deduced  from  the 
enlargement  of  the  cylinder.  The  fibres  or 
layers  between  the  limits  r\  and  r,  being  com- 
pressed, will  be  diminished  in  thickness.  The 
compression  of  a  piece  an  inch  in  thickness  by  a  unit  stress 


Fig.  85 


208  STRUCTURAL  MECHANICS. 

will  be  p-^-E,  §  10,  and  of  one  dr  thick  will  be  pdr+E.  The 
total  diminution  of  thickness  between  r\  and  r,  from  what  it  was 

at  first,  will  therefore  be  -^  I    pdr. 

But  the  annular    fibre  or  ring  whose  radius  is  r  and  length 
27zr  has  been  elongated  t-^E  per  inch  of  length.      Its  length  will 

now  be  2^(1+-^)  and  its  radius  rl  i  +-g  )•  The  internal  radius 
must  similarly  have  become  ri(I+^)>  where  /  is  the  value  of  / 
for  radius  r\.  The  thickness  r— TI  has  now  become  r(i  +  — ) 
— ril  i  +  — j,  and,  by  subtracting  this  value  from  r—r^  there  is 

found  the  diminution  of  thickness,  ri-L—r-^.     This  expression 

may  be  equated  with  the  previous  one  for  decrease  of  thickness, 
or 

.  /    * 


Since  the  first  term  is  constant,   there  is  now  obtained  by 
differentiating  this  equation, 

-d(tr)=pdr,       or       tdr+rdt  +  pdr=o.     ...     (2) 

Add  (i)  and  (2),  and  multiply  by  r  to  make  a  complete  differ- 
ential.    Then  integrate 

2(t  +  p)rdr+r2(dt+dp)  =o; 
=  constant;       /.   =r!2(/  +  ^)  =r22(/'  +  />2).    .     (3) 


Again,  subtract  (i)  from  (2),  and  then  integrate 

dt—dp=o.    t—  ^>=constant;     /.   =}—pi=}'—p2-    •     (4) 
From  (3)  and  (4)  are  obtained,  by  addition  and  subtraction, 
f-pi.rf  f+pl  l-pir 


ENVELOPES.  209 

If  the  internal  radius  is  given,  the  external  radius,  and  hence 
the  required  thickness,  r2-ri,  is  found  by  eliminating  f  from 
(3)  and  (4), 


If  p2  is  atmospheric  pressure,  it  may  be  neglected  when  pi 
is  large.     In  that  case 


As  r2  becomes  infinite  when  the  denominator  of  (6)  is  zero, 
it  appears  that  no  thickness  will  suffice  to  bring  /  within  the 
safe  unit  stress,  if  pi  exceeds  }  +  2p2. 

These  formulas  do  not  apply  to  bursting  pressures,  nor  to 
those  which  bring  /  above  the  elastic  limit;  for  E  will  not  then 
be  constant.  They  serve  for  designing  or  testing  safe  construc- 
tion. 

Examples.  —  Cylinders  of  the  hydraulic  jacks,  for  forcing  forward 
the  shield  used  in  constructing  the  Port  Huron  tunnel,  were  of  cast 
steel,  12  in.  outside  diameter,  8  in.  diameter  of  piston,  with  J  in.  clear- 
ance around  same;  pressure  2,000  Ib.  per  sq.  in. 

r22     36-16     /+  2,000 

;?=W=F^     /=    3°' 

A  cast-iron  water-pipe  at  the  Comstock  mine  was  6  in.  bore, 
2j  in.  thick,  and  was  under  a  water  pressure  of  1,500  Ib.  on  the  sq.  in., 
or  about  3,400  ft.  of  water.  Here  /=2,7yo  Ib.  per  sq.  in.  for  static 
pressure,  while  the  formula  for  a  thin  cylinder  gives  1,800  Ib. 

199.  Greater  Pressure  on  Outside.  —  In  this  case  the  direc- 
tion or  sign  of  /  will  be  reversed,  it  being  compression  in  place 
of  tension.  From  the  preceding  equations,  without  independent 
analysis,  by  making  t  negative,  there  result: 

_  d(pr)  =  -tdr\    d(tr)  =  pdr. 
pdr  +  rdp—tdr=o;    tdr+rdt—pdr=o. 


210  STRUCTURAL  MECHANICS. 

The  outer  radius  and  pressure  will  now  be  taken  as  given 
quantities,  and  the  unit  compression  in  the  ring  at  any  point 
will  be 


(8) 


which  becomes,  if  pi  is  neglected  as  small, 


The  external  pressure  p2  must  be  less  than  J(/+#i)>  if  f\  is 
to  have  any  value.  It  will  be  seen  from  t  in  (7)  that  the  com- 
pression is  greatest  at  the  interior. 

Example.  —  An  iron  cylinder  3  ft.  internal  diameter  resists  1,150 
Ib.  per  sq.  in.  external  pressure.  The  required  thickness,  if  /=  9,000 
lb.,  is  given  by 


f2=2o.Q  in.     Thickness  =  3  in. 

200.  Action  of  Hoops. — To  counteract  in  a  greater  or  less 
degree  the  unequal  distribution  of  the  tension  in  thick  hollow 
cylinders  for  withstanding  great  internal  pressures,  hoops  are 
shrunk  on  to  the  cylinders,  sometimes  one  on  another,  so  that 
before  the  internal  pressure  is  applied,  the  internal  cylinder  is 
in  a  state  of  circumferential  compression,  and  the  exterior  hoop 
in  a  state  of  tension.  If  the  internal  pressure  on  the  hoop  is 
computed,  for  a  given  value  of  /  in  the  hoop,  and  this  pressure 
is  then  used  for  p2  on  the  cylinder,  the  allowable  internal  pressure 
pi  on  the  cylinder  consistent  with  a  permissible  /  in  this  cylinder 
can  be  found.  There  is,  however,  an  uncertainty  as  to  the 
pressure  p2  exerted  by  the  hoop. 


ENVELOPES.  211 

Examples,  —  A  hoop  i  in.  thick  is  shrunk  on  a  cylinder  of  6  in. 
external  radius  and  3  in.  internal  radius,  so  that  the  maximum  unit 
tension  in  the  hoop  is  10,000  Ib.  per  sq.  in.  This  stress,  by  §  198, 
will  be  due  to  an  internal  pressure  on  the  hoop  of  1,530  Ib.  per  sq.  in. 


For  7  =  6  -  or      49,10.000+ft. 

\J  \io,ooo  —  _/>i/  36      10,000—^1 

This  external  pressure  p2  on  the  cylinder  will  cause  a  compressive  unit 
stress  in  the  interior  circumference  of  the  cylinder  when  empty,  after 
the  hoop  is  shrunk  on,  of  4,080  Ib.,  and  will  permit  an  internal  pressure 
in  the  bore  of  8,448  Ibs.  per  sq.  in.,  consistent  with  /=  10,000  Ib.  For 

,        ,  .     The  cylinder  alone,  without  the  hoop,  would 
9       10,000—  pi  +  3,060 

allow  a  value  of  pi  given  by  —  =  —  -  —   —  —  ,  or  pi  =  6,000  Ib.     If  the 

9       10,000—  pi 

cylinder  had  been  4  in.  thick,  the  internal  pressure  might  have  been 
6,900  Ib.  The  gain  with  the  hoop,  for  the  same  quantity  of  material, 
is  1,548  Ib.,  or  some  22%. 

Hydraulic  cylinder  for  a  canal  lift  at  La  Louviere,  Belgium,  6  ft. 
9  in.  interior  diameter,  4  in.  thick,  of  cast  iron,  hooped  with  steel. 
Hoops  2  in.  thick  and  continuous.  When  tested,  before  hooping,  one 
burst  with  an  internal  pressure  of  2,175  H).  Per  scl-  m->  one  at  2,280  Ib., 
and  a  third  at  2,190  Ib.  These  results,  if  the  formula  is  supposed  to 
apply  at  rupture,  give  an  average  tensile  strength  of  23,400  Ib.  per  sq. 
in.  The  hoops  were  supposed  to  have  such  shrinkage  that  an  internal 
pressure  of  540  Ib.  per  sq.  in.  would  give  a  tension  on  the  cast  iron  of 
1,400  Ib.,  and  on  the  steel  of  10,600  Ib.  per  sq.  in.  The  ram  is  6  ft. 
6f  in.  diam.  and  3  in.  thick,  of  cast  iron,  an  example  of  the  greater 
pressure  outside. 

201.  Thick  Hollow  Sphere.  —  Greater  pressure  on  inside.  Let 
Fig.  85  represent  a  meridian  section  of  the  sphere.  Suppose  /,  /, 
etc.,  to  be  perpendicular  to  the  plane  of  the  paper.  The  entire 
normal  pressure  on  the  circle  of  radius  ri  will  be  pixri2,  and  the 
tension  on  the  ring  between  radii  r\  and  r  will  be  n(p\r-f—  pr2). 
Any  ring  of  radius  r  and  thickness  dr  will  carry  2nrtdr,  and  hence 
is  derived  the  first  equation 

n(piri2  -pr2)  =  2n  frt  dr,     or     -d(pr2)  =  2rt  dr. 

Ur-L 


dr=o. 

The  second  equation  will  be  the  same  as  obtained  for  the  cylinder. 
-d(tr)  =  pdr,     or    rdt  +  tdr  +  pdr  =  o. 


212  STRUCTURAL  MECHANICS. 

Strike  out  the  common  factor  r  from  the  first  equation,  multiply 
the  second  by  2,  and  subtract. 

2rdt—rdp=o,     or     2dt—dp=o. 
.'.    2t—p  =  constant;     .'.   =  2J—pi  =  2J'—p2.       .     .     (9) 

Again,  add  the  first  to  the  second  and  multiply  by  r2. 


=o. 


constant;    /.   =  fi3(/  +  #i)  =  r23(}'+p2).    .   (10) 
From  (9)  and  (10), 


2f-pl     rj  /  +  li.  2)-p 


. 

v  "~~~  I          o  *  js  |~  2       o 

3 


I          o  *  js  |~  o  • 

3         ^       3  3  r3       3 


<-' 


These  formulas  are  not  applicable  to  bursting  pressures  for 
the  reason  given  before.  For  a  finite  value  of  r%,  p\  must  be  less 
than  2/4-3^2-  If  p2  is  atmospheric  pressure,  it  may  be  neglected, 
and 


(]±PJ\ 
'•I -Pi  )' 


202.  Sphere:    Greater  Pressure   on   Outside. — Here  again  t 
changes  to  compression  or  reverses  in  sign,  yielding 


+  r3'     3      '    f         3  r-       3 

^T^) (13) 


That  ri  shall  be  greater  than  zero  requires  that 

203.  Diagrams  of  Stress. — Curves  may  be  drawn  to  represent 
the  variation  of  p  and  t  in  the  four  preceding  cases.  They  are 
all  hyperbolic,  and,  if  r  is  laid  off  from  the  centre  O  on  the  hori- 
zontal axis,  each  curve  will  have  the  vertical  axis  through  O 


ENVELOPES. 


213 


for  one  asymptote,  and  for  the  other  a  line  parallel  to  the  hori- 
zontal axis,  at  a  distance  indicated  by  the  first  term  in  each  value 
of  /  or  p.  The  four  accompanying  sketches  show  the  various 
curves.  The  values  of  /  and  /',  the  unit  stresses  in  the  material 
at  the  interior  and  exterior,  which  correspond  to  the  given  values 
of  pi  and  p2,  are  found  at  the  extremities  of  the  abscissas  which 
represent  r\  and  r^.  The  error  which  would  arise  from  con- 
sidering /  as  uniformly  distributed  is  manifest.  The  dotted 


Fig.  86 


\x    Fig.  87 


*Vty 


\F-ig.89 


circles  show  the  respective  cylinders  or  spheres.  Fig.  86  gives 
the  external  and  internal  tensile  stress  for  pi  in  the  interior  of  a 
thick  cylinder.  Fig.  87  shows  the  distribution  of  compression 
when  the  greater  pressure  is  from  without.  Figs.  88  and  89 
represent  thick  spheres  under  similar  pressures. 

204.  Tank  with  Conical  Bottom. — A  water-tank  of  radius  r 
may  be  built  with  a  conical  bottom  and  be  supported  only  at  the 
perimeter  by  a  circular  girder.  The  pressure  of  the  water  in 
pounds  per  square  inch  at  any  point  is  p= 0.434  X  depth  of  point 
below  surface.  In  the  cylinder  the  stress  per  unit  of  length  on 
a  vertical  joint  is  pr\  the  stress  on  a  horizontal  joint  is  equal 
to  the  weight  of  the  sides  lying  above  the  joint  and  is  generally 
insignificant. 


214 


STRUCTURAL  MECHANICS. 


Any  horizontal  joint  in  the  cone  such  as  A  A  of  Fig.  90  must 
carry  a  load,  W,  composed  of  the  weight  of  water  in  the  cylinder 
whose  base  is  the  circle  A  A,  the  water  in  the  cone  ABA  and 
the  weight  of  the  metal  shell  of  the  same  cone.  The  last  item 
is  comparatively  insignificant.  As  the  cone,  like  the  sides,  is 
built  of  thin  plates,  the  stresses  in  the  cone  must  always  act 
tangentially  to  the  shell,  so  W  '  -t-2nri  is  the  vertical  component 
of  T\t  the  stress  per  unit  of  length  on  the  horizontal  joint,  and 

WsecO 


which  varies  from  zero  at  B  to  a  maximum  at  C. 


To  find  the  stress  on  a  joint  along  an  element  of  the  cone 
imagine  a  ring  of  slant  height  unity  to  be  cut  out  by  two  hori- 
zontal planes  as  shown  in  the  figure.  Substitute  for  the  pressure, 
p,  which  acts  normally  around  the  ring,  the  two  components, 
p  tan  0  and  p  sec  6.  Of  these  the  former  acting  along  the  ele- 


ENVELOPES.  215 

ments  causes  no  stress  on  an  elemental  joint  while  the  latter 
causes  a  tension  in  the  ring  of 

T2  =  pri  sec  6  =  pp. 

It  can  be  proved  that  ri  sec  6  is  the  radius  of  curvature  of  the  conic 
section  cut  out  by  a  plane  normal  to  the  element,  B  C,  hence  T% 
is  equal  to  the  pressure  into  the  radius  of  curvature  as  shown 
in  §  195.  This  tensile  stress  varies  from  zero  at  B  to  a  maxi- 
mum at  C. 

The  total  weight  of  the  tank  and  contents  is  carried  by 
a  circular  girder,  which  in  turn  is  supported  by  three  or  more 
posts,  consequently  the  girder  is  subjected  to  both  bending 
and  torsional  moments.  At  C  the  stress,  7\,  is  resolved  into 
vertical  and  horizontal  components,  the  former  of  which  is 
W  -T-  2r.r  and  is  the  vertical  load  per  unit  of  length  of  girder ;  the 

TFtantf 
latter  is which  causes  a  compressive  ring  stress  in  the 

r  W  tan  0 
girder  of . 

27T 

Example.— A  circular  tank,  40  ft.  in  diameter,  has  a  conical  bottom 
for  which  #=45°.  The  depth  of  water  above  the  apex  is  60  ft.  Weight 
of  cu.  ft.  of  water,  62.5  Ib.  Sec  0=1.414.  Tension  in  lowest  vertical 
ring  of  sides  is  40X0.434X20X12  =  4,170  Ib.  per  lin.  in.  Tension  in 
radial  joint  of  cone  at  A,  half  way  up,  is  50X0.434X120X1.414  =  3,680 
Ib.  per  lin.  ir .  of  joint.  Same  at  C  is  4oXo.434X  240 X  1.414=  5,900  Ib. 
per  lin.  in. 

For  tension  on  horizontal  joint  at  A,  half  way  up,  W=^X 
100(50  +  ^X10)62.5.  ^1  =  ^X10X53.33X62. 5X1.414=23,600  Ib.  per 
lin.  ft.  =  1,960  Ib.  per  lin.  in.  of  joint.  At  C,  7^  =  ^X20X46.67X62. 5X 
1.414=41,250  Ib.  per  ft.  =  3,440  Ib.  per  lin.  in.  The  vertical  com- 
ponent of  TI  at  C  is  41,250-1-1.414=29,200  Ib.  per  ft.  of  girder.  As 
the  horizontal  and  vertical  components  are  equal,  the  compression  in 
the  girder  is  29,200X20=584,000  Ib. 

205.  Tank  with  Spherical  Bottom.— The  stresses  in  the  spheri- 
cal bottom  are  found  in  the  same  way  as  in  a  conical  bottom. 
Any  horizontal  joint  as  A  A  carries  the  cylinder  of  water  whose 
base  is  the  circle  A  A,  the  segment  of  water  ABA,  and  that  part 


216 


STRUCTURAL  MECHANICS. 


of  the  shell  below  the  joint.     The  volume  of  the  segment  is 
— Ja)  if  a  =  rf(i—cosO).     If  the  weight  on  the  whole  joint 
is  W,  the  vertical  component  of  the 
stress    per    linear    unit    of    joint    is 
W  -T-  2nf  sin  0  and  the  stress  is 

T!= — r-^2. 

27ir  sm^0 

The  tension  per  unit  of  length  on 
any  meridian  joint  is,  by  §  196, 


Fig.  91 

pressive  stress  of 


W 


The  vertical  load  on  the  girder  is 
W  +  27:r  per  unit  of  length,  and  the 
horizontal  force  applied  to  the  girder 

W 

by  the  bottom  of  the  tank  is  - 

27zr  tan  a 

per  unit  of  length,  which  causes  a  corn- 
in  the  girder. 


27r  tan  a 

The  stresses  in  the  spherical  bottom  are  smaller  than  those 
in  the  conical  bottom. 

Example. — A  circular  tank,  40  ft.  diameter  and  40  ft.  high,  has  a 
spherical  bottom  for  which  a =45°.  Then  ^=28.3  ft.  and  the  ex- 
treme height  =  48. 3  ft.  Tension  in  radial  joint  at  bottom  =  ^X62.5X 
28.3X48.3  =  42,645  Ib.  per  linear  ft.  =  3, 554  Ib.  per  in.  of  joint.  Ten- 
sion in  radial  joint  at  A,  half  way  up,  where  0=22^°,  is  40,770  Ib.  per 
ft.  or  3,397  Ib.  per  in.  of  joint.  At  C,  tension  is  35,350  Ib.  per  ft.,  or 
2,946  Ib.  per  in.  of  radial  joint.  Tension  in  horizontal  joint  at  A 
is  41,760  Ib.  per  ft.,  or  3,480  Ib.  per  in.,  and  at  C  is  39,220  Ib.  per  ft., 
or  3,268  Ib.  per  in.  of  joint.  Compression  in  circular  girder =  554,700  Ib. 

206.  Conical  Piston. — The  cone  C  B  C  of  Fig.  92  represents 
a  conical  piston  of  radius  r,  subtending  an  angle  20,  with  a  normal 
steam  pressure,  p,  per  unit  of  area  applied  over  its  exterior  or 
interior,  the  supporting  force  being  supplied  by  the  piston-rod 
at  B.  The  force  in  the  direction  of  the  rod  on  any  section  A  A 
of  radius  r\  is  px(r2  —ri2)  which  becomes  at  the  vertex  pxr2,  the 
force  on  the  piston-rod.  This  force  will  be  compression  on  the 
rod  and  tension  in  the  cone,  if  p  acts  on  the  exfe-ior  of  the  cone, 


ENVELOPES. 


217 


and  the  reverse  if  p  acts  within  the  cone.     The  unit  stress  in  the 

metal  of  the  cone  at  this  section  will  be  found  by  multiplying 

this  force  by  sec  6,  and  then  dividing 

by  the  cross-section,  2nrit,  in  which  /  is 

the  thickness  of  the  metal.     The  unit 

stress  on  the  circumferential  section  is 

then 

P 

^l       2T\l  Fig.  92 

which  is  a  maximum  at  the  piston-rod  if  t  is  constant. 
The  unit  stress  at  A  on  a  radial  section  is,  by  §  204, 

PP      pTl 

p2=—=—rsec  6. 
t        t 

When  pi  is  compressive,  p2  is  tensile  and  vice  versa. 

Example.— Conical  piston,  Fig.  92.  ^=24  in.,  radius  of  rod=3  in., 
0=69°.  Thickness  for  PI  =  17  in.  is  1.5  in.;  for  7-1  =  8  in.  is  1.9  in. 
Steam  pressure  =  100  Ib.  per  sq.  in.  Sec  #=2.79.  For  /i=  17  in.,  pi  — 
ioo(242-i72)  2.79^(2X17X1. 5)  =  i, 570  Ib.  per  sq.  in.;  p2= 
100X17X2.79-^1. 5  =  3,160 Ib.persq. in.  For  ri  =  8  in.,  ^1  =  4,700  Ib. 
and  p2=i,i7$  Ib.  For  alternating  stresses  on  steel  castings  these 
stresses  are  satisfactory.  See  example  §  175. 


Fig.  93 


207.  Dome. — A  dome  subjected  to  vertical  forces  symmetri- 
cally placed  around  its  axis,  such  as  its  own  weight,  may  be  treated 
as  follows :  C  B  C  of  Fig.  93  represents  a  meridian  section  of  a 


218  STRUCTURAL  MECHANICS. 

dome,  a  hemisphere  as  shown,  but  the  results  to  be  deduced 
are  true  for  any  surface  of  revolution  about  a  vertical  axis.  If 
a  horizontal  plane,  A  A,  is  passed  through  the  dome  to  cut  out  a 
circle  of  radius  r±  and  all  the  weight  from  the  crown  to  that  sec- 
tion is  denoted  by  W,  the  stress  in  the  shell  per  linear  unit  of 
circumferential  joint  is 

Wcscd 


TI  is  always  compressive  and  is  a  maximum  at  the  base. 

To  find  the  stress  on  a  meridian  section  pass  two  horizontal 
planes  through  the  dome  so  as  to  cut  out  a  thin  ring  of  mean 
radius  r±.  If  the  total  load  above  the  ring  is  Wi  and  the  total 
vertical  force  supporting  the  ring  is  W2,  the  weight  of  the  ring 
is  Wz—Wi.  As  the  shell  of  the  dome  is  thin  the  stresses  in  the 
shell  are  tangential  to  the  surface  and  the  ring  is  acted  upon 
by  a  system  of  forces  around  its  circumference  as  shown  on  the 
right  side  of  the  figure.  Resolve  the  forces  into  vertical  and 
horizontal  components  as  shown  on  the  left.  Acting  upon  the 
upper  edge  per  unit  of  mean  length  of  ring  is  the  vertical  com- 

Wi  Wi  ctn  6l 

ponent  -  and  the  horizontal  component  -  .      By  sub- 

stituting W2  and  62  for  W\  and  Q\  the  components  acting  on  the 
lower  edge  are  found.  The  vertical  components,  together  with 
the  weight  of  the  ring  balance  among  themselves,  but  there  is 

an  unbalanced  horizontal  force  of  H=  -  (W\  ctn  #1  —  W2  ctn  62) 

27CT\ 

which  causes  tension  or  compression  in  the  ring  depending  on 
whether  it  acts  outward  or  inward.  The  stress  in  the  ring  is, 
therefore,  Hr\  and  its  intensity  per  linear  unit  of  joint,  T2,  is 
found  by  dividing  by  the  width  of  the  ring  measured  along  the 
meridian.  At  the  crown  T2  is  compressive  and  equal  to  TI, 
but  it  diminishes  as  A  is  taken  lower  and  lower  down  and  becomes 
tensile  in  the  lower  part  of  the  structure. 

208.  Resistance  of  a  Ring  to  a  Single  Load.  —  When  a  ring  is 
acted  upon  by  two  equal  and  opposite  forces  as  shown  in  Fig.  94, 
the  curve  becomes  flatter  at  A  and  sharper  at  B,  showing  that 


ENVELOPES. 


219 


bending  moments  of  opposite  sign  are  set  up  at  these  points. 
From  conditions  of  symmetry  it  is  seen  that  each  quadrant  is 
acted  upon  by  vertical  forces  of  %W  together  with  an  unknown 
moment  at  each  end.  Imagine  the  quadrant  to  be  removed  from 


w 


Fig.  94 


the  circle  and  horizontal  levers  to  be  attached  at  A  and  B  so  that 
the  forces  can  be  moved  horizontally  such  a  distance  as  to  cause 
the  actual  moments  existing  at  A  and  B.  For  equilibrium  the 
forces,  when  moved,  must  be  applied  in  the  same  line,  and  the 
moment  at  any  section  of  the  ring  will  be  determined  when  the 
line  of  application  of  the  forces  is  fixed. 

To  fix  that  line  the  deformation  of  the  arc  must  be  considered. 

By  §  88,  — =~pj  wm'cn  becomes  da=^jds  if  the  angle  between 

the  two  radii  of  Fig.  44  is  da.  This  equation  gives  the  change 
in  the  angle  between  two  right  sections  ds  apart,  caused  by  the 
bending  moment,  and  is  true  for  curved  beams  as  well  as  straight. 
To  find  the  change  between  the  two  sections  a  distance  5  apart  inte- 
grate from  zero  to  s.  In  the  ring  under  consideration  the  tan- 
gents at  A  and  B  remain  horizontal  and  vertical  respectively, 
as  seen  from  the  condition  of  symmetry,  hence  the  change  in  the 
angle  between  right  sections  at  A  and  B  is  zero  and  as  E  and  / 

/J*r 
Mds=o.    M  at  any  point  C  is  JPFXDC  and 
. 

must  be  expressed  in  terms  of  r  and  6  to  be  integrated.     Then 


7 

i 

Jo 


(a-rcos6)d0=o; 


220  STRUCTURAL  MECHANICS 


[2 
(ad— r  sin  6)  =o  =  Ja^— r; 
o 


2T 

a=—  =0.6366^. 


The  bending  moments  in  the  hoop  can  now  be  determined. 

If  a  ring  is  acted  upon  by  four  normal  forces  90°  apart  as  is 
the  horizontal  girder  around  the  base  of  a  water  tank  when  the 
posts  are  inclined,  the  bending  moments  at  various  points  of  the 
girder  may  be  found  for  each  pair  of  forces  independently  and 
the  results  added  algebraically. 

Examples. — i.  What  is  the  net  thickness  required  for  a  boiler  shell 
60  in.  in  diameter  to  carry  120  Ib.  steam  pressure?  What  the  gross 
thickness  allowing  for  riveting,  and  the  size  and  pitch  of  rivets  ? 

f  in. ;  T9F  in. ;  J  in. ;  3  rows,  2|  in.  pitch. 

2.  What  weight  applied  at  top  of  circumference  and  resisted  at 
bottom  ought  a  cast-iron  pipe,  12  in.  diam.,  J  in.  thick,  and  6  ft.  long, 
to  safely  carry,  if  /=  12,000  Ib.  ? 

3.  Determine  the  thickness  of  the  cast  iron  cylinder  of  a  ro-ton  hy- 
draulic jack  to  work  under  a  pressure  of  i.ooo  Ib.  per  sq.  in.  if  /=  4,000  Ib 


CHAPTER  XIV. 
PLATE    GIRDERS. 

209.  I  Beam.  —  A  rolled  beam  of  I  section  may  be  considered 
composed  approximately  of  three  rectangles,  —  two  flanges,  each 
of  area  S*,  and  a  web  of  area  Sw.  The  depth  between  centres 
of  stress  of  the  flange  sections  may  be  denoted  by  &',  which  is 
also  very  nearly  the  depth  of  the  web.  Then  the  resisting  moment 
of  the  two  flanges  will  be  fSfh',  and  that  of  the  web,  since  M  for 
a  rectangle  is  IJbh2,  is  j-\Swh'.  The  value  for  the  entire  section 
will  be 


Hence  comes  the  rule  that  one-sixth  of  the  web  may  be  added 
to  one  flange  area  in  computing  the  resisting  moment  of  an  I 
beam.  The  extreme  depth  of  the  beam  ought  not,  however, 
to  be  used  for  hf.  The  approximate  distance  between  centres. 
of  gravity  of  the  flanges  will  answer,  since  it  is  a  little  short  of 
the  true  value  for  the  flanges  and  a  little  longer  than  is  correct  for 
the  web. 

210.  Plate  Girder.  —  A  portion  of  a  plate  girder  and  a  section 
of  the  same  is  shown  in  Fig.  95.  Such  a  structure  acts  as  a  beam 
and  is  designed  to  resist  the  maximum  bending  moments  and 
shears  to  which  it  may  be  liable.  It  may  be  loaded  on  top,  or 
through  transverse  beams  connected  to  its  web.  It  is  used  when 
the  ordinary  sizes  of  I  beams  are  not  strong  enough  to  resist 
the  maximum  bending  moment.  As  the  flanges  may  be  varied 
in  section  by  the  use  of  plates  where  needed,  as  shown  at  the 
right,  there  may  be  more  economy  of  material  in  using  a  built 
beam  rather  than  a  rolled  one,  if  the  required  maximum  section 
is  large. 


221 


222 


STRUCTURAL  MECHANICS. 


4    4- 

t    4 

i    4 


i-      4 


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f       H 


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4- 

4 

O 

4 

4 

4 

4 

4 

t 

1- 

-t 

4 

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-t 

+ 

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-ij 

4 

!  !  {  i   !  i  i    !    1   i  y=LL| 

4 
4 

4 

4  j| 

4 

4 

4 

4, 

4 

4 

4 

-(- 

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J 

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;t 
4 

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1111111     1     i     I    1    i   -»-  4- 

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g 


PLATE  GIRDERS.  223 

The  web  A  is  made  of  sufficient  section  to  resist  the  maxi- 
mum shear,  and  the  rest  of  the  material  is  thrown  into  the  flanges, 
B,  where  it  will  be  farthest  removed  from  the  neutral  axis  and 
hence  most  efficient  in  resisting  bending  moment.  As  the  thick- 
ness of  the  web  plate  is  usually  restricted  to  one-fourth  inch,  and 
in  girders  of  any  magnitude  to  three-eighths  inch,  as  a  minimum, 
it  appears  that  the  material  in  the  web  practically  increases  with 
the  depth  of  the  girder.  As  the  stress  in  either  flange  multiplied 
by  the  distance  between  the  centres  of  gravity  of  flanges  resists 
the  bending  moment,  the  material  in  the  flanges  decreases  as  the 
depth  increases;  hence  the  most  economical  depth  is  that  which 
makes  the  total  material  in  the  web,  including  the  splice  plates 
and  stiffeners,  as  near  as  may  be  equal  to  that  in  the  two  flanges. 
Depth  of  beam  contributes  greatly  to  stiffness,  when  a  small 
deflection  is  particularly  desirable,  and  the  depth  may  in  such 
a  case  be  so  great  as  to  make  the  web  the  heavier. 

211.  Web. — Some  engineers  apply  the  rule  of  §  209  to  a  plate 
girder  and  consider  the  flange  section  to  be  increased  by  one- 
eighth  of  the  web  section,  the  fraction  one- eighth   being    used 
instead  of  one-sixth  because  of  the  weakening  of  the  web  by  rivet- 
holes  ;  but  the  more  commonly  received  practice  is  to  consider  the 
flanges  alone   as  resisting  the  bending  moment  at  any  section 
and  the  web  as  carrying  all  the  shear  uniformly  distributed  over 
its  cross-section  as  shown  in   §  73.     If  the  web  is  required  to 
resist  its  share  of  the  bending  moment,  web  splices   must  be 
designed  to  transmit  the  stresses  due  to  bending  as  well  as  those 
due  to  shear.     Although  web  plates  are  generally  made  of  the 
thinnest  metal  allowable,  the  designer  is  prevented  from  using 
too  thin  a  web  plate  by  the  fact  that  thin  plates  offer  so  little 
bearing  resistance  to  the  rivets.     If  the  plate  is  too  thin  it  will 
generally  be  impossible  to  put  in  a  sufficient  number  of  rivets 
to  connect  the  web  to  the  flange. 

212.  Flanges. — If  the  maximum  bending  moments  are  com- 
puted for  a  number  of  points  in  the  span  of  the  girder,  they  can 
be  divided  by  the  allowable  unit  tensile  stress  times  the  effective 
depth  (that  is,  the  distance  between  centres  of  gravity  of  flanges) 
tr  give  the  required  net  sections  of  the  tension  flange  at  those 


224  STRUCTURAL   MECHANICS. 

points.  The  compression  flange  is  usually  made  of  the  same 
gross  section  as  the  tension  flange.  The  deduction  for  rivet-holes 
in  the  latter,  which  is  not  necessary  in  the  former,  compensates 
for  the  slightly  lower  unit  stress  allowed  for  compression.  The 
compression  flange  must  be  wide  enough  not  to  bend  sideways 
like  a  strut  between  the  points  at  which  it  is  stayed  laterally.  As 
the  web  checks  such  lateral  flexure  in  some  degree,  and  as  the 
flange  stress  varies  from  point  to  point,  it  is  impracticable  to 
apply  a  column  formula.  In  practice  the  unsupported  length  of 
plate  is  not  allowed  to  exceed  a  certain  number  of  times  the  width 
of  the  flange:  for  railroad  bridges  different  specifications  give 
from  12  to  20;  for  girders  carrying  steady  load  only,  as  high  a 
number  as  30  may  be  used. 

The  selection  of  the  plates  and  angles  to  give  the  required 
flange  section  is  largely  a  matter  of  judgment.  The  angles  must 
be  large  enough  to  support  well  the  compression  flange-plates 
and  to  be  able  to  transmit  the  increments  of  stress  from  the  web 
to  such  flange-plates.  Hence  the  area  of  the  flange-angles  should 
be  a  considerable  portion  of  the  total  flange  section.  For  railway 
girders  it  is  often  specified  that  the  section  of  the  angles  shall 
be  at  least  one-half  that  of  the  whole  flange,  or  that  the  largest 
angles  procurable  shall  be  used. 

213.  Length  of  Flange-plates. — As  the  bending  moment  varies 
from  point  to  point  in  a  girder,  the  required  flange  section  will 
vary  in  the  same  way,  being,  in  general,  greatest  at  the  half  span. 
If  the  section  at  that  point  is  made  up  of  two  angles  with  flange- 
plates,  the  girder  may  be  made  to  approximate  to  a  beam  of 
uniform  strength  by  running  the  flange-plates  only  such  distance 
from  the  centre  as  they  are  needed. 

Inspection  of  the  necessary  sections  will  show  how  far  from 
the  two  ends  of  the  girder,  as  I  K,  the  flange-angles,  with  rivet- 
holes  deducted,  will  suffice  for  the  required  flange  section.  From 
K  to  the  corresponding  distance  from  the  other  abutment  the  first 
plate  must  extend.  A  reasonable  thickness  being  used  for  that 
plate,  with  a  deduction  for  rivet-holes  in  the  tension  flange,  it 
can  again  be  seen  where  a  second  plate  will  be  needed,  if  at  all. 
This  determination  can  be  neatly  made  on  a  diagram  of  maxi- 


PLATE  GIRDERS. 


22$ 


mum  moments,  or  on  the  similar  diagram  showing  the  necessary 
sections.  Extend  the  plate  either  way  a  small  additional  dis- 
tance, to  relieve  the  angles  and  assure  the  distribution  of  stress 
to  the  plate.  The  thicker  plate,  if  there  is  any  difference  in  thick- 
ness, should  be  placed  next  to  the  angles. 

When   the   girder  carries    a   uniformly  distributed    load   the 


£ 


Fig.  £6 


Fig.  97 


required  flange  areas  vary  as  the  ordinates  to  a  parabola  (Fig.  96) 
and  the  length  of  any  flange-plate  is  given  by 


*  /vF 

#-*Va> 


in  which  /  is  the  length  of  the  girder  from  centre  to  centre  of 
base-plates,  S  the  net  flange  area  required  at  the  centre,  and  s 
the  net  area  of  the  plate  whose  length  is  desired  plus  the  area 
of  such  plates  as  may  lie  outside  it. 

If  the  girder  is  so  long  that  the  plates  or  angles  must  be 
spliced,  additional  cross-section  must  be  supplied  by  covers  at 
the  splices,  with  lengths  permitting  sufficient  rivets  to  transmit 
the  force.  Even  compression  joints,  though  milled  and  butted 
together,  are  spliced  in  good  practice.  The  net  area  of  the  cover- 
plate  and  splice  angles  should  be  equal  to  that  of  the  largest 
piece  spliced.  Only  one  piece  should  be  cut  at  any  one  section, 
and  enough  lap  should  be  given  for  the  use  of  sufficient  rivets 
to  carry  the  stress  the  piece  would  have  carried  if  uncut. 

214.  Rivet  Pitch. — If  a  strip  of  web  (Fig.  97)  of  a  width 
equal  to  the  pitch  of  the  rivets  connecting  the  flanges  to  the  web 
is  cut  out  by  two  imaginary  planes,  shearing  forces,  F,  act  on  the 
two  sides  forming  a  couple  with  an  arm  equal  to  the  pitch.  Under 
the  usual  assumption  that  the  web  carries  shear  only,  there  are 


226  STRUCTURAL  MECHANICS. 

no  other  forces  acting  on  the  planes  of  section,  and  the  only  forces 
which  can  keep  the  strip  from  rotating  are  those  supplied  by  the 
two  flange  rivets.  Hence  by  equating  the  two  couples  the  pitch 
can  be  found. 

rivet  value  X  depth 
shear 

As  the  flange-angles  are  supposed  to  be  fully  stressed  at  the 
point  where  the  first  flange-plate  begins,  the  increments  of  flange 
stress  coming  out  of  the  web  must  pass  through  the  flange-angles 
and  into  the  flange-plate.  The  rivets  connecting  the  flange-angles 
and  the  flange-plates  must  therefore  resist  the  same  stress  as  do 
the  rivets  connecting  the  web  plate  and  the  flange-angles,  and  the 
above  formula  applies  to  both,  although  the  rivet  values  in  the 
two  cases  will  be  different.  The  rivets  through  the  web  are  in 
bearing  (or  double  shear  if  the  web  is  thick),  while  those  in  the 
flange-plate  are  in  single  shear  and  occur  in  pairs.  But  practi- 
cally the  pitch  in  one  leg  of  an  angle  must  be  the  same  or  an  even 
multiple  of  the  pitch  in  the  other,  so  that  the  rivets  may  be  stag- 
gered. 

Make  the  pitch  of  rivets  in  inches  and  eighths,  not  decimals; 
do  not  vary  the  pitch  frequently,  and  do  not  exceed  a  six-inch 
pitch,  so  that  the  parts  may  be  kept  in  contact.  If  flange-plates 
are  wide,  and  two  or  more  are  superimposed,  another  row  of 
rivets  on  each  side,  with  long  pitch,  may  be  required  to  insure 
contact  at  edges.  Care  must  be  taken  that  a  local  heavy  load 
at  any  point  on  the  flange  does  not  bring  more  shear  or  bearing 
stress  on  rivets  in  the  vertical  legs  of  the  flange-angles  than  allowed 
in  combination  with  the  existing  stress  from  the  web  at  that  place 
and  time. 

Webs  are  occasionally  doubled,  making  box  girders,  suitable 
for  extremely  heavy  loads.  The  interior,  if  not  then  accessible 
for  painting,  should  be  thoroughly  coated  before  assembling. 

If  the  web  must  be  spliced,  use  a  splice  plate  on  each  side 
for  that  purpose,  having  the  proper  thickness  for  rivet  bearing 
and  enough  rivets  to  carry  all  the  shear  at  that  section;  there 
should  be  two  rows  of  rivets  on  each  side  of  the  joint. 


PLATE   GIRDERS.  227 

215.  Stiffeners. — At  points  where  a  heavy  load  is  concentrated 
on  the  girder,  stiffeners,  C,  consisting  of  an  angle  on  each  side, 
should  be  riveted  to  throw  the  load  into  the  web  and  to  prevent 
the  crushing  of  the  girder.  They  should,  for  a  similar  reason, 
be  used  at  both  points  of  support,  D.  Such  stiffeners  act  as 
columns  and  may  be  so  figured,  but  as  the  stress  in  them  varies 
from  a  maximum  at  one  end  to  zero  at  the  other,  a  good  rule 
is  to  consider  the  length  of  the  column  equal  to  half  the  depth 
of  the  girder. 

Since  the  thrust  at  45°  to  the  horizontal  tends  to  buckle  the 
web,  and  the  equal  tension  at  right  angles  to  the  thrust  opposes 
the  buckling,  it  is  conceivable  that  a  deep,  thin  web,  while  it 
has  more  ability  to  carry  such  thrust  as  a  column  or  strut  than  it 
would  have  if  the  tension  were  not  restraining  it,  may  still  buckle 
under  the  compressive  stress ;  and  it  is  a  question  whether  stiffeners 
may  not  be  needed  to  counteract  such  tendency.  They  might 
be  placed  in  the  line  of  thrust,  sloping  up  at  45°  from  either 
abutment,  but  such  an  arrangement  is  never  used.  They  are 
placed  vertically,  as  at  C,  and  spaced  by  a  more  or  less  arbitrary 
rule. 

A  common  formula  is :  The  web  of  the  girder  must  be  stiffened 
if  the  shear  per  square  inch  exceeds 

d 

10,000  -75y, 

where  d  =  clear  distance  between  flange-angles,  or  between  stifieners 
if  needed,  and  /  =  thickness  of  web.  Another  rule  calls  for 
stiffeners  at  distances  apart  not  greater  than  the  depth  of  the 
girder,  when  the  thickness  of  the  web  is  less  than  one-sixtieth 
of  the  unsupported  distance  between  flange-angles.  There  is 
no  rational  method  of  determining  the  size  of  stiffeners  used 
only  to  keep  the  web  from  buckling.  The  usual  practice  is  about 
this :  make  the  outstanding  leg  of  the  angle  3 J  inches  for  girders 
less  than  4  feet  deep,  4  inches  for  girders  5  feet  deep,  and  5  inches 
for  girders  over  7  feet  deep. 

Experience  appears  to  show  that  stiffeners  are  not  needed  at 
such    frequent    intervals    as    the   formula    would    demand.     An 


228  STRUCTURAL  MECHANICS. 

insufficient  allowance  for  the  action  of  tension  in  the  web  in 
keeping  the  compression  from  buckling  it,  is  probably  the  cause 
of  the  disagreement. 

Interior  stiffeners  may  be  crimped  at  the  ends,  or  fillers  may 
be  used  under  them  to  avoid  the  offset.  End  stiffeners  and 
stiffeners  under  concentrated  loads  should  not  be  crimped;  they 
should  fit  tightly  under  the  flange  that  the  load  may  pass  in  at 
the  ends. 

Example. — A  plate  girder  of  30  ft.  span,  load  3,000  Ib.  per  ft., 
/=  15,000  Ib.  per  sq.  in.,  PF=9o,ooo  Ib.,  and  Mmax=\Wl 
=  4,050,000  in.-lb.  Assume  extreme  depth  as  42  in.,  effective  depth, 
39  in.  Net  flange  section  at  middle  =  4,050,000 -r-  (39- 15,000)  =  7  sq.  in. 
A  f-in.  web  42  in.  deep  will  have  15!  sq.  in.  area.  Two  flanges, 
each  7  sq.  in.  net + allowance  for  rivet-holes,  will  fairly  equal  the  web. 
Use  j-in.  rivets. 

Let  the  flange-angles  be  2  —  4X 3  X f  in.  =  4.96  sq.  in.  Deduct  2  holes, 
|Xf=o.66.  Net  plate  =7  — 4.3  =  2.7  sq.  in.  A  plate  9Xf=3f  sq.  in.; 
deduct  two  holes  =  0.66,  leaving  2.71  sq.  in.  Two  angles  and  plate, 
gross  section=4.96+3.37  =  8.33  sq.  in.  Resisting  moment  of  net 
section  of  angles  =  4.3><  15,000X3.25=209,625  ft.-lb.  Such  a  bending 
moment  will  be  found  at  a  distance  x  from  either  end,  given  by  P\x— 
i -3, 000^=209,625.  #=5.9  ft.  .'.  Cut  off  the  plate  5  ft.  from  each 
end. 

Shearing  value  of  one  f-in.  rivet  at  10,000  Ib.  per  sq.  in.  =  4,400  Ib. 
Bearing  value  in  f-in.  plate  at  20,000  lb.=  5,6oo  Ib.  Max.  shear  in 
web =45,000  Ib.  Pitch  for  flange-angles,  since  bearing  resistance  is 
less  than  double  shear,  =  5,600- 39 -^-45,000=4.85=4}  in.  Make  3-in. 
pitch  for  2  ft.,  then  4|-in.  for  6  ft.,  then  6-in.  pitch  to  middle.  Rivets 
in  end  web  stiffeners,  45,000-7-5,600=9.  Max.  shear  in  web  =  45,000 

-*- (42  •§)  =  2,860;  10,000— 75— =  2,800,  since  ^=42  — 6,  and  stiff- 
eners are  needed. 


CHAPTER  XV. 
SPRINGS  AND    PLATES. 

216.  Elliptic  Spring. — The  elliptic  spring  is  treated  in  §  97. 
If  the  load  on  the  spring  is  W  and  the  span  is  /,  the  deflection  is 


i  Wl3 
v=- 

The  fibre  stress  is 


V     16  Eld 


and  the  work  done  upon  the  spring  is  %Wv,  which  is  equal  to  the 
resilience  of  the  spring.     Hence 


if2 

Resilience  =  —  "       =  —  -=  •  volume. 
32  EI0     6  E 

217.  Straight  Spring. — If  a  beam  of  uniform  section,  fixed  at 
one  end,  has  a  couple  or  moment  applied 
to  it  (Fig.  98)  in  place  of  a  single  transverse 
force,  it  will,  as  shown  in  §  89,  bend  to  the 
arc  of  a  circle.     The  deflection  will  be,  if   ''          Eig'98 
/  is  the  length  of  the  beam, 

Ml2       fl2 

V=~W7=- 
2EI 

The  work  done  by  the  rotation  of  a  couple  is  the  product 

of  its  moment  by  the  angle  through  which  it  turns.     For  a  deflec- 

229 


230  STRUCTURAL  MECHANICS. 

tion,  dv,  the  free  end  of  the  spring  turns  through  an  angle  2dv  -^-/; 
hence 

rMj      4E7  fv  J       2EI  0       //    /2 

Resilience  =  2  /  —rdv = ~^T-  I    vdv  =  -j^-v2  = 5  •  -p. 

I         L  /•*       /  /<*  9A/i^    /* 

t/O  t/O  •'/I       •*-' 

For  a  rectangular  section  bh,  these  quantities  become 

6M12     }l2 
'"'Ebh^EK 

•L    I2  i    }2 

Resilience  =  T-  •  -F  •  ^  =  T '  ~F< '  volume. 
o    Jit  o    Hi 

For  a  circular  section  the  number  6  in  the  last  expression 
will  be  replaced  by  8. 

218.  Coiled    Spring. — In   practice   the   rectangular   or   cylin- 
drical tar  is  bent  into  a  spiral  and  subjected  to  a  couple  which, 
as  a  couple  can  be  rotated  in  its  plane  without  change,  acts  equally 
at  all  sections  of  the  spring.     The  developed  length  of  the  spiral 
is/. 

219.  Helical  Spring. — A  cylindrical  bar  whose  length  is  /  and 
diameter  d,  when  fixed  at  one  end  and  subjected  to  a  twisting 
moment  T  =  Pa  at  the  other,  if  the  elastic  limit  is  not  exceeded, 

32  Tl 

by  §  84,  is  twisted  through  an  angle  6=    „„.    The  work  ex- 
pended in  the  torsion  is 


From  §  84,  0=^%-,  and  therefore 
c# 

i    qi2  7td2J     i    q? 

Resilience  =  — -^ / = -^  •  volume. 

4    C     4        4    C 

fC\ 

If  C=—E  and  qi=—  /,  work= ^-volume,  while  for  flexure, 

D  j  j 

i    /2 
as  shown  in  the  preceding  section,  work  =  jr'~j?'  volume,  a  smaller 


SPRINGS  AND   PLATES.  231 

quantity;  hence  a  spring  of  given  weight  can  store  more  energy 
if  the  stresses  are  torsional  than  if  they  are  bending. 

If  this  bar  is  bent  into  a  helix  of  radius,  0,  and  the  force,  P, 
is  applied  at  the  centre  in  the  direction  of  the  axis  of  the  cylinder, 
the  moment,  Pa,  will  twist  the  bar  throughout  its  length.  Then 

i6Pa  nd* 

=  °r        ^' 


The  deflection  of  the  spring  is  v  =  a0,  since,  as  the  force  P  de- 
scends, the  spring  decends,  and  the  action  is  the  same  as  if  the 
spring  remained  in  place  and  the  arm  revolved  through  an  angle 
d.  The  force  P  is  too  small  to  cause  any  appreciable  compression 
(or  extension)  of  the  material  in  the  direction  of  its  length. 

32  Pa2l     ial  qi     ^.xna2  q1 
= 


if  n  =  number  of  turns  of  the  helix  and  /  =  znan. 

If  the  section  of  the  spring  is  not  circular,  substitute  the 
proper  value  of  q\  or  the  resisting  moment  from  §  85.     If  the 

(d'*\ 
i  —  -7J-  )  .     For  a 

square  section  and  a  given  deflection,  P  will  be  about  65  per  cent. 
of  the  load  for  an  equal  circular  section.     C  for  steel  is  from 

10,500,000  to  12,000,000. 

Example.  —  A  helical  spring,  of  round  steel  rod,   i  in.  diameter, 
making  8  turns  of  3-in.  radius,  carries  1,000  Ib. 
i6-i,ooo-v7  4'  22-8-Q-  i 


22  7.1-12,000,000 


=1.1    in. 


220.  Circular  Plates.  —  The  analysis  of  plates  supported  or 
built  in  and  restrained  at  their  edges,  and  loaded  centrally  or 
over  the  entire  surface,  is  extremely  difficult.  The  following 
formulas  from  Grashof's  "Theorie  der  Elasticitat  und  Festig- 
keit  "  may  be  used.  The  coefficient  of  lateral  contraction  is 
taken  as  •  J,  or  m  =  4. 


232  STRUCTURAL  MECHANICS. 

I.  Circular  plate  of  radius  r  and  thickness  /,  supported  around 
its  perimeter  and  loaded  with  w  per  square  inch. 

}x  =  unit  stress  on  extreme  fibre  in  the  direction  of  the  radius, 

at  a  distance  x  from  the  centre  ; 

/2,=unit  stress  perpendicular  to  the  radius,  in  the  plane  of  the 
plate,  at  the  same  distance  x  from  the  centre. 


117  wr2  189  wr* 


For  the  same  value  of  /,  the  maximum  stress  is  independent 
of  r,  provided  the  total  load  wnr2  is  constant. 

II    Same  plate,  built  in  or  fixed  at  the  perimeter. 


At  the  centre,  }x  =  iw     At  the  circumference  }y  is  zero,  and  fx 
is  maximum. 

45  wr2  45  wr4 

*xmax-     64    P'      ^°    256£/3' 

III.  Circular  plate  supported  at  the  perimeter  and  carrying 
a  single  weight  W  at  the  centre.     Loaded  portion  has  a  radius  TQ. 


These  expressions  become  maxima  for  x=ro,  and  the  second  is 
the  greater. 

117  Wr2 
V°    64^^' 

For  values  of  r+r0  =  10,     20,     30,     40,     50,     60, 

/was.=i-4     !-7     i-9    2-°     2-J     2-2     W  +  P. 


SPRINGS  AND   PLATES.  233 

[f  r0=o,  the  stress  becomes  infinite,  as  is  to  be  expected, 
sh.ce  W  will  then  be  concentrated  at  a  point,  and  the  unit  load 
be»,  omes  infinitely  great.  It  is  not  well  to  make  rQ  very  small. 

IV.  Same  plate,  built  in  or  fixed  at  the  perimeter. 

4$W.f 


The  maximum  value  of  /  is  }yj  for  X  =  TQ. 

45  Wr2 


For  yalues  of  r-^0  =  io,     20,     30,     40,     50,     60, 

}max,  =  1.0     1.3     1.5     1.6     1.7     1.8    W+P. 

221.  Rectangular  Plates.  —  The  problem  .of  the  resistance  of 
rectangular  plates  is  more  complex  than  that  of  circular  plates. 
Grashof  gives  the  following  results  : 

V  Rectangular  plate  of  length  a,  breadth  b,  and  thickness  /, 
a  >  by  built  in  or  fixed  at  edges  and  carrying  a  uniform  load  of 
w  pei  square  inch. 


The  most  severe  stress  occurs  at  the  centre  in  the  direction 
that  isj  on  a  section  parallel  to  a.     If 


wa 


w 
The  deflection  at  the  centre  is  VQ=   ,  ,  ,  4  —  =rz,  and  for  a 

*  6 

wa 
square  plate, 


VI.  Plate  carrying  a  uniform  load  of  w  per  square  inch  and 
supported  at  rows  of  points  making  squares  of  side  a.  Fire- 
box sht^t  with  stay  bolts. 

i$wa2  15  wa4 


234  STRUCTURAL  MECHANICS. 

Navier  gives  formulas  for  rectangular  plates  which  are  sup- 
posed to  be  very  thin.  Approximate  values  from  those  formulas 
are  as  follows  : 

VII.  Rectangular  plate,  as  in  V,  but  supported  around  the 
edges. 

a4b2     iv  a4b*       w 


VIII.  Rectangular  plate,  supported  at  edges  and  carrying  a 
single  weight  W  at  centre. 


a*b      W  aW      W 

>  =  2'2 


For  the  same  total  load,  /  is  independent  of  the  size  of  the 
plate,  provided  the  ratio  a  to  b  and  the  thickness  are  unchanged. 

Example.  —  A  steel  plate  36  in.  square  and  J  in.  thick,  supported  at 
edges,  carries  430  Ib.  per  sq.ft.,  or  3  Ib.  per  sq.  in.  /=  0.92  -£-36-  36-3-16 
=  14,300  Ib. 

0.19     364-3-43       ,. 
v=  ---  =  *  in. 
4     30,000,000 


CHAPTER  XVI. 
REINFORCED   CONCRETE. 

222.  Reinforced  Concrete. — As  concrete  has  small  tensile 
strength  and  is  likely  to  crack  when  built  in  large  masses,  it  can 
be  reinforced  to  advantage  by  steel  bars  or  wire  netting  imbedded 
within  it.  This  form  of  construction  is  much  used  and  is  espe- 
cially applicable  to  beams  and  slabs,  in  which  the  steel  is  placed 
near  the  tension  edge.  By  its  use  a  great  saving  of  material 
in  masonry  structures  can  often  be  effected,  since  the  strength 
of  the  structure  can  be  depended  upon,  rather  than  its  weight. 
Among  its  advantages  for  buildings  may  be  mentioned  the  fact 
that  it  is  fireproof  and  that  the  metal  is  protected  from  rust. 
The  expansion  and  contraction  of  the  two  materials  from  changes 
of  temperature  are  so  nearly  alike  that  heat  and  cold  produce 
no  ill  effects.  When  used  in  beams  of  any  considerable  span, 
there  is  the  disadvantage  that  the  dead  load  is  a  large  proportion 
of  the  total  load. 

To  compute  the  strength  of  a  structure  composed  of  two 
materials  which  act  together,  the  modulus  of  elasticity  of  each 
material  must  be  known,  and  it  is  here  that  the  chief  difficulty 
in  computing  the  strength  of  reinforced  concrete  members  lies. 
The  modulus  of  elasticity  of  concrete  is  uncertain;  it  not  only 
depends  upon  the  composition  of  the  concrete,  but,  for  a  given 
concrete,  varies  with  the  age,  while  for  a  particular  specimen  the 
stress  deformation  diagram  is  not  a  straight  line  as  for  steel,  but 
the  ratio  of  the  stress  to  the  deformation  decreases  with  the  load. 
The  concrete  takes  a  permanent  set,  even  for  small  loads,  in 
consequence  of  which  a  reinforced-concrete  beam,  after  being 
released  from  a  load,  is  in  a  condition  of  internal  stress. 

235 


236  STRUCTURAL  MECHANICS. 

Considering,  then,  our  lack  of  knowledge  of  the  exact  stresses 
to  be  expected  in  the  concrete,  it  is  best  to  adopt  as  simple  a 
method  of  computation  as  is  reasonable. 

223.  Beams. — Beams  may  be  figured  according  to  the  common 
theory  of  flexure,  and  the  following  assumptions  will  be  made: 

1.  The  steel  and  the  surrounding  concrete  stretch  equally. 

2.  Cross-sections,  plane  before  bending,  are  plane  after  bending, 

3.  The  modulus  of  elasticity  is  constant  within  the  working 
stress. 

4.  The  tension  is  borne  entirely  by  the  steel.     As  the  tensile 
strength  of  concrete  is  low  and  as  a  crack  in  the  beam  would 
entirely  prevent  the    concrete   from  resisting  tension,  this   is  a 
proper  assumption  to  make  and  is  on  the  side  of  safety. 

5.  There  is  no  initial  stress  on  the  section. 


<  b—± 

Neutral 

*          — 

i 

Axis 

y 

F;>-  90 


From  these  assumptions  it  follows  that  the  stresses  acting 
on  the  cross-section  are  as  shown  in  Fig.  99. 
Let  Es  =  modulus  of  elasticity  of  steel; 
Ec=       "        "        "        "  concrete; 
f  —  F  —  F  • 

C,          J^/g     .    J_jc , 

}s  =  unit  stress  in  steel ; 

}c=  "       "      "  concrete  at  extreme  compression  fibre; 

A  =  deformation  of  fibre  at  unit  distance  from  the  neutral 

axis,  between  two  sections  originally  unit  distance 

apart ; 

kbh= sectional  area  of  steel; 
y= distance  from  neutral  axis  to  extreme  compression 

fibre; 
c  =  y  +  h; 
n  =  a  numerical  coefficient. 


REINFORCED   CONCRETE.  237 

By  §  10, 


or  the  ratio  between  the  unit  stresses  is 

ja  _E.h-y       h-y 

T.-R.  y      '~T 

As  the  total  tension  on  the  cross-section  must  equal  the  total 
compression,  for  a  rectangular  cross-section 


(2) 

Combining  (i)  and  (2)  gives 


fc     2kh     '  y 
Solving  this  equation  for  y  gives  the  location  of  the  neutral  axis, 

-ch (3) 


This  equation  shows  that  for  beams  made  of  a  given  quality 
of  concrete  the  location  of  the  neutral  axis  depends  only  upon 
the  percentage  of  reinforcement. 

The  resultant  of  the  compressive  stresses  on  the  cross-section 
is  applied  at  a  point  \y  from  the  top  of  the  beam  and  together 
with  the  tensile  stress  in  the  steel  forms  a  couple  whose  arm  is 
h—^y.  The  moment  of  this  couple,  which  is  the  resisting  moment 
of  the  beam,  is 


(4) 


the  first  expression  giving  the  moment  in  terms  of  the  stress  in 
the  concrete  and  the  second  in  terms  of  the  stress  in  the  steel. 
If  the  value  of  y  from  (3)  is  substituted  in  (4)  there  results 

.....    (5) 


238 


STRUCTURAL  MECHANICS. 


As  long  as  /c  and  js  are  the  unit  stresses  in  the  concrete  and  in 
the  steel,  the  two  forms  of  (5)  must  be  equal,  but  if  fc  and  }8  are 
arbitrarily  chosen  working  stresses,  that  form  of  (5)  must  be  used 
in  designing  which  will  give  the  smaller  value  of  M  .  Therefore 


use  nc  when  nc)c<nsfs  or  when  ^>  —,. 


The  actual  fibre  stresses, 


however,  can  be  made  to  assume  any  given  ratio  by  changing  the 
proportion  of  steel  reinforcement,  and  that  proportion  can  be 
found  by  eliminating  y  from  (i)  and  (2)  with  the  result 


, 
2 


(6) 


Equation  (5)  can  be  readily  applied  to  the  design  of  beams 
by  tabulating  the  values  of  nc  and  ns  for  different  values  of  k. 


e=8. 

e—  10. 

e=  12. 

c 

n0 

Ht 

c 

tie 

Ms 

c 

He 

ns 

.003 

.  196 

.092 

.00280 

.217 

.  101 

.00278 

•235 

.I08 

.00277 

.004 

.223 

.103 

.00370 

.246 

•113 

.00367 

.265 

.121 

.00368 

.005 

.246 

•ii3 

.00459 

.270 

.123 

•00455 

.291 

•J31 

.00452 

.006 

.266 

.  121 

.00546 

.292 

.132 

.00542 

•314 

.140 

•°°537 

.007 

.284 

.128 

.00633 

•3ii 

.140 

.00628 

•334 

.148 

.00622 

.008 

.300 

•135 

.OO72O 

.328 

.  146 

.00713 

•353 

•155 

.00706 

.010 

•328 

.  146 

.00891 

.358 

•157 

.00880 

.384 

.167 

.00872 

.012 

•353 

•ISS 

.384 

.I67 

.411 

.177 

.OI4 

•375 

.164 

.407 

.I76 

•435 

.186 

.Ol6 

•394 

.171 

.428 

.183 

•457 

.194 

.018 

.411 

.177 

•  447 

.190 

•476 

.200 

.020 

.428 

.183 

•  463 

.  196 

•493 

.206 

When  reinforced  concrete  beams  are  tested  to  destruction,  they 
sometimes  fail  by  the  opening  of  diagonal  cracks  which  seem  to 
follow  in  a  general  way  the  lines  of  principal  stress.  To  prevent 
that  mode  of  failure  most  designers  provide  some  form  of  rein- 
forcement in  a  vertical  plane.  Stirrups  or  loops  of  wire  lying 
in  planes  of  section  and  spaced  at  intervals  less  than  the  depth 
of  the  beam  are  often  employed,  but  no  satisfactory  method  of 
determining  their  size  and  spacing  has  been  proposed. 


REINFORCED  CONCRETE.  239 

Another  mode  of  failure  to  be  guarded  against  is  the  slipping 
of  the  steel  in  the  surrounding  concrete.  That  the  beam  may  not 
be  weak  in  this  respect,  the  change  of  stress  in  the  bars,  between 
two  sections  a  short  distance  apart  (say  one  inch),  must  not 
exceed  the  area  of  the  surface  of  the  bars  between  the  two  sections 
multiplied  by  the  safe  unit  adhesive  stress  of  concrete  to  steel. 
This  change  of  stress  between  two  sections  one  inch  apart  is 
F+h(i—  Jc),  since  the  shear  in  the  beam  measures  the  change 
of  bending  moment  at  any  section,  as  is  shown  in  §  56.  Rods 
which  have  been  roughened  or  corrugated  may  be  used  to  diminish 
the  likelihood  of  slipping. 

Example. — Design  a  beam  of  16  ft.  span  to  carry  an  external  load 
of  500  lb.  per  ft.  if  }c=$oo  lb.,  fs=  16,000  lb.,  and  e=io,  assuming 
the  beam  to  weigh  300  lb.  per  ft.  M  =J- 800- 16-16-12= 307, 200 
in.-lb.  If  each  material  is  to  be  stressed  to  its  limit,  the  proportion  of 
reinforcement  needed  is  IO-T-  2 -32  -42  =  0.00372,  or  say  0.4%.  Then 
»c=o.ii3  and  bh2= 307, 200-7- (0.113X500)  =  5,430.  b=i2%  in.,  h=2i 
in.,  and  the  area  of  steel  is  0.004X12^X21  =  1.03  sq.  in.,  or  say  4— T9f 
in.  rounds.  If  there  are  2  in.  of  concrete  below  the  rods,  the  beam 
weighs  150X121x23-7-144=284  lb.  per  ft. 

If  0.8%  of  steel  is  used,  instead  of  0.4%,  bh2= 4210.  &=iojin.,  h=2o 
in.,  and  1.68  sq.  in.  of  steel  is  needed ;  use  i  —  f  in.  and  2  —  }  in.  rounds. 

Adhesion  between  concrete  and  steel  F  max.  =  800X8  =  6,400  lb. 
For  first  beam  6,400-=-  2 1(1  —  ^X0.246)  =33 2  lb.  Superficial  area  of 
4— •&  rounds=4-IV-2T2-==7-°7  sq.  in.  332-^7-o7  =  47  lb.  per  sq.  in. 
For  second  beam  359^7.86=46  lb.  per  sq.  in. 

224.  Columns. — Reinforced  concrete  columns  are  built  with 
steel  rods  embedded  parallel  to  and  spaced  symmetrically  about 
the  axis.  The  rods  should  be  tied  together  by  wire  or  bands  at 
intervals  not  greater  than  the  diameter  of  the  column.  A  common 
design  is  a  square  section  with  the  rods  near  the  corners.  As 
the  ratio  of  length  to  breadth  is  generally  small,  it  is  usual  to 
design  such  columns  as  short  blocks. 

The  ratio  between  the  intensity  of  the  stress  in  the  steel  and 
in  the  concrete  is 


240  STRUCTURAL  MECHANICS. 

and  the  total  load  the  column  may  carry  is 


in  which  S  is  the  total  area  of  cross-section  of  the  column. 

With  the  usual  working  stresses  and  ratio  of  E8  to  Ec  employed 
in  designing,  the  greatest  allowed  unit  stress  in  the  reinforcement 
is  so  small  that  economy  requires  the  amount  of  steel  to  be 
reduced  as  much  as  reasonable,  hence  the  load  P  will  be  but 
slightly  greater  than  fcS. 

225.  Safe  Working  Stresses.  —  The  following  safe  unit  stresses 
may  be  used  in  buildings: 

Concrete,  bending  .......................        500  Ib.  per  sq.  in. 

'  *          direct  compression  .............        350  '  '     '  l    lt    " 

Steel,  tension  ...........................   16,000  "    "    "    " 

Adhesion  of  concrete  to  steel  ..............          50  '  l    "    "    li 

Es+Ec  =  io  is  a  fair  average  value;  the  ratio  may  vary  con- 
siderably without  materially  affecting  the  proportions  of  the 
beam. 

The  weight  of  reinforced  concrete  is  about  150  Ib.  per  cu.  ft. 

The  concrete  should  be  rich;  one  part  cement,  two,  sand; 
four,  broken  stone  is  a  good  mixture,  the  stone  being  broken  to 
pass  through  a  f-inch  ring.  It  should  be  mixed  wet  and  placed 
with  great  care  to  insure  the  proper  bedding  of  the  steel. 

The  proportion  of  steel  reinforcement  to  use  in  beams  is  a 
question  of  economy,  which  is  most  easily  solved  by  designing  a 
number  of  sections  of  the  same  ratio  of  b  to  h,  but  with  different 
values  of  k,  and  figuring  the  cost  per  foot  of  each.  A  considerable 
variation  in  k  affects  the  cost  but  slightly,  for  ordinary  prices  of 
concrete  and  of  steel.  The  percentage  of  reinforcement  is  usually 
between  \  and  ij;  £=0.007  ig  an  average  value. 

The  concrete  lying  below  the  reinforcing  rods  serves  merely 
to  protect  the  steel.  For  fireproofing  two  inches  is  sufficient. 
Steel  thoroughly  covered  with  concrete  does  not  rust,  and  bars, 
which  were  covered  with  rust  when  placed  in  concrete,  have  been 
found  to  be  bright  when  removed  after  some  time. 


INDEX. 


Angle  of  repose,  181 
Annealing,  29 
Ashlar,  30 
Axis,  neutral,  56,  6 1,  79 

Bauschinger's  experiments,  154 
Beams,  2,  51,  87,  107 

bending  moments,  41,  44 

cantilever,  45 

Clapeyron's  formula,  118 

column  and  beam,  149 

continuous,  116 

curved,  60 

deflection,  87 

elastic  curve,  87 

fixed,  107 

flexure  of,  87 

flitched,  102 

I  beams,  69,  173,  221 

impact,  103 

inclined,  60 

modulus  of  rupture,  59 

moving  loads  on,  48 

neutral  axis,  56 

oblique  loading,  78 

reactions,  38 

reinforced  concrete,  236 

resilience  of,  103 

restrained,  107 

sandwich,  102 

shaft  and  beam,  84 

shear,  external,  43 

shear,  internal,  65 

slope,  88 

stiffness,  89 

stresses  in,  54,  65,  179 

three-moment  theorem,  Il6,  122 

tie  and  beam,  130 

timber,  68 

torsion  on,  84 

uniform  strength,  63,  97 

work,  internal,  104 


Bending  and  compression,  149 

and  tension,  130 

and  torsion,  84 

Bending  moment,  see  Moment,  41 
Bessemer  process,  25 
Blocks  in  compression,  15,  137 
Boilers,  189,  203 

rivets,  198 

working  stresses,  166 
Bricks,  31 
Bridges,  shear  in  panel,  52 

working  stresses,  162,  164 
Buildings,  working  stresses,  165 
Burnettizing,  21 

Cantilevers,  45,  47 
Carbon  and  iron,  22 
Cast  iron,  22 

properties  of,  23 

working  stresses,  1 66 
Cement,  33 
Centrifugal  force,  133 
Clapeyron's  formula,  118 
Clay,  31 

Coefficient,  see  Modulus. 
Columns,  2,  137 

beam  and  column,  149 

deflection,  142 

designing,  146 

eccentric  load,  148 

ends,  fixed  or  hinged,  146 

Euler's  formula,  142 

flexure,  direction  of,  141,  146 

Gordon's  formula,  144 

ideal  column,  142 

lacing-bars,  151 

pin  ends,  147 

radius  of  gyration,  146 

Rankine's  formula,  144 

reinforced  concrete,  239 

short,  137,  145 

straight-line  formula,  147 

241 


242 


INDEX. 


Columns,  swelled,  148 

timber,  162 

transverse  force  on,  149 

working  stresses,  162 

yield-point,  141 
Combined  stresses,  3 

bending  and  compression,  149 
"  "    tension,  130 

"  "    torsion,  84 

tension  and  torsion   133 
Compression,  3,  15,  137 

bending  and,  149 

eccentric  load,  137 

granular  materials  under,  15 
Concrete,  35,  235 
Concrete-steel,  235 
Cone,  stresses  in,  213,  216 
Connecting-rod,  132 
Continuous  beams,  116 
Cooper's  lines,  190 
Crank,  84 

Curvature  of  beams,  88 
Curve,  elastic,  87 

stress-deformation,  9 
Cylinders,  thick,  206 

thin,  203 

Deflection  of  beams,  87 

simple,  89 

restrained,  107 

uniform  strength,  97 
Deflection  of  columns,  142 
Deformation,  6,  184 
Distortion,  7,  186 
Dome,  217 
Ductility,  6,  1 7,  26,  29 

Earth  pressure,  181 
Eccentric  load,  127,  137,  148 
Elastic  curve,  87 
Elastic  limit,  9,  154 
Elasticity,  modulus  of,  £,  6 

cast  iron,  23 

concrete,  235,  240 

steel,  26 

stone,  30 

timber,  22 

wrought  iron,  24 
Elasticity,   shearing  modulus  of,   C,  7, 

186 

Ellipse  of  stress,  174 
Elongation,  work  of,  8,  10,  13,  27 
Envelopes,  203 
Equilibrium,  conditions  of,  I 
Euler's  formula,  142 
Eyebars,  131,  134 

Fatigue  of  metals,  154 
Flexure,  common  theory  of,  54 


Girders,  see  Beams. 

Girder,  plate,  see  Plate  girder,  221 

Gyration,  radius  of,  71 

Hooks,  129 
Hoops,  210 

I  beam,  69,  173,  221 

Impact,  159 

Inertia,  moment  of,  57,  71 

product  of,  76 
Iron,  cast,  22 

malleable,  28 

wrought,  24 

Joints,  masonry,  138 
riveted,  195 

Lattice  bars,  15 1 

Launhardt-Weyrauch  formula,  156 

Lime,  32 

Lines  of  principal  stress,  180 

Linseed  oil,  36 

Loads,  dead  and  live,  159 

eccentric,  127,  137,  148 

sudden  application,  14 

wheel  loads  on  beam,  49 

Machinery,  working  stresses,  166 
Manganese  in  steel,  26 
Masonry,  30 

working  stresses,  1 66 
Materials,  19 
Middle  third,  138 
Modulus  of  elasticity,  see  Elasticity,  6 

of  resilience,  13 

of  rigidity,  186 

of  rupture,  59 

section,  57 
Moment,  bending,  41 

maximum,  44 

on  pins,  200 

position  of  load  for  maximum,  49 

sign  of,  41 

Moment  of  inertia,  7 1 
Moment  of  resistance,  57 

oblique  loading,  78 
Moment,  torsional,  81 
Mortar,  32 

Neutral  axis,  56,  61,  79 
Nuts,  134 

Oblique  load  on  beam,  78 
Open-hearth  process,  25 

Paint,  36 
Parallel  rod,  133 
Pedestals,  1 66 
Permanent  set,  8 


INDEX. 


243 


Phosphorus  in  steel,  26 
Pier-moment  coefficients,  121 
Pig  iron,  23 
Pins,  109,  200 

distribution  of  shear  in,  67 

friction,  147 

working  stresses,  163 
Pipes,  203 

Piston,  conical,  189,  216 
Plaster,  33 
Plate  girder.  221 

stresses  in  web.  69,  173 
Plates,  resistance  of,  231 
Polar  moment  of  inertia,  71 
Portland  cement,  34 
Posts,  see  Columns,  137 
Power,  shaft  to  transmit,  83 
Principal  stresses,  170 

lines  of,  180 
Product  of  inertia,  76 
Puddling  furnace,  24 
Pull  and  thrust  at  right  angles,  172 
Punching  steel,  effect  of,  28 

Radius  of  gyration,  71 

Rafter,  60 

Rankine's  column  formula,  144 

Rankine's  theory  of  earth  pressure,  181 

Reactions  of  beams,  38 

Rectangular  beams,  58 

Repose,  angle  of,  181 

Resilience,  definition,  13 

modulus  of,  13,  103 

of  bar,  13 

of  beam,  103 

of  springs,  229 

Resisting  moment,  see  Moment,  57 
Restrained  beams,  107 
Retaining  wall,  182 
Rigidity,  modulus  of,  187 
Ring  under  normal  pressure,  205 

under  single  load,  218 
Rivets,  192 

plate  girder,  225 

steel  for,  26 

working  stresses,  163 
Rollers,  163 
Rubble,  31 
Rupture,  modulus  of,  59 

Safe  working  stresses,  153 
Sandwich  beams,  102 
Screw  threads,  134 
Secondary  stresses,  158,  197 
Section  modulus,  57 
Set,  permanent,  8 
Setting  of  cement,  34 
Shafts,  8 1 

working  stresses,  166 
Shear,  3,  43 


Shear,  deflection  due  to,  105 

derivative  of  bending  moment,  45 

distribution  on  section  of  beam,  65 

modulus  of  elasticity,  C,  186 

position  of  load  for  maximum,  48,  50 

sign  of,  43 

timber  beams,  68 

two  shears  at  right  angles,  169,  172, 
186 

work  of,  105 
Shearing  planes,  176 
Sign  of  bending  moment,  41 

compression  and  tension,  4 

shear,  43 
Silicon  in  iron,  23 
Slope  of  beam,  88 
Spangenberg's  experiments,  153 
Sphere,  stresses  in,  205,  211 
Splices,  195 

in  plate  girder,  223,  225,  226 
Springs,  98,  229 
Steel,  25 

shearing  and  punching,  28 

structural,  26 

tool,  28 

working  stresses,  162 
Steel  concrete,  235 
Stiffness  of  beams,  89 
Stiffners,  227 
Stirrups,  238 
Stone,  29 

Straight- line  formula,  147 
Strain,  see  Deformation. 
Strength,  beams  of  uniform,  63,  97 

cross- sections  of  equal,  62 

ultimate,  n 
Strength  of  cast  iron,  23 

steel,  26 

timber,  22,  160 

wrought  iron,  24 
Stresses,  2,  167 

alternating,  154 

conjugate,  170 

distribution  on  section  of  beam,  54, 

65,  179 

ellipse  of.  174 

internal,  2,  167 

lines  of  principal,  180 

principal,  170 

reversal  of,  155,  162 

secondary,  158,  197 

sign  of,  3 

unit,  4,  167 

working,  160,  240 
Stress- deformation  diagram,  9 
Struts,  see  Columns,  2,  137 
Sudden  loading,  effect  of,  14 
Sulphur  in  steel,  26 

Tanks,  213 


244 


INDEX. 


Tempering,  26,  28 
Tension,  3,  127 

bending  and,  130 

connections,  133,  196 

eccentric  load,  127 

torsion  and,  128,  133 
Three-moment  theorem,  116,  122 
Tie,  2,  127 

and  beam,  130 
Timber,  19 

column  formulas,  161 

modulus  of  elasticity,  22 

shear  in  beams,  68 

strength  of,  22 

working  stresses,  160 
Torsion,  81 

bending  and,  84 

resilience  of,  230 

tension  and,  133 

twist  of  shaft,  83 
Trees,  growth  of,  19 
Twist  of  shaft,  83 

Ultimate  strength,  n 


Uniform  strength,  beams  of,  63,  97 
Unit  stresses,  4 

Varnish,  36 

Varying  cross-section,  12 

Volume,  change  of,  183 

Wall,  retaining,  182 

middle  third,  138 
Web  of  plate  girder,  223 

stresses  in,  69,  173 
Welding,  26 
Wheel  loads.  49 
Wohler's  experiments,  153 
Wood,  19 
Work  of  elongation,  8,  10,  13.  27 

flexure,  104 

shear,  105 

springs,  229 
Working  stresses,  153 

reinforced  concrete,  240 
Wrought  iron,  24 

Yield-point,  10 


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DAY  AND  TO  $1.OO  ON  THE  SEVENTH  DAY 
OVERDUE. 


NOV   II 


!3ttW 


APR 


8    194? 

1943 
1943 


LIBRARY  USE 

RETURN  TO  DESK  FROM  WHICH  BORROWED 

LOAN  DEPT. 

THIS  BOOK  IS  DUE  BEFORE  CLOSING  TIME 
ON  LAST  DATE  STAMPED  BELOW 


US£ 

Ri 

M2  '64-12.1 


LD  62A-50m-2,'64 
(E3494slO)9412A 


General  Library 

University  of  California 

Berkeley 


